Solution of quadratic equations and Nature of roots Questions in English

(B) Given the cubic equation: $x^3-6x^2+6x-5=0$.
To transform the roots by increasing them by $h$,we replace $x$ with $(x+h)$.
The new equation becomes: $(x+h)^3-6(x+h)^2+6(x+h)-5=0$.
Expanding the terms: $(x^3+3x^2h+3xh^2+h^3)-6(x^2+2xh+h^2)+6(x+h)-5=0$.
Grouping the $x^2$ terms: $x^3 + (3h-6)x^2 + (3h^2-12h+6)x + (h^3-6h^2+6h-5) = 0$.
Since the new equation does not contain the $x^2$ term,the coefficient of $x^2$ must be zero: $3h-6=0$.
Solving for $h$: $3h=6 \Rightarrow h=2$.

(D) First,factorize the given equation $x^4-10x^3+37x^2-60x+36=0$.
By testing values,we find $x=2$ and $x=3$ are roots.
Dividing by $(x-2)^2(x-3)^2$,we get $(x-2)^2(x-3)^2=0$.
Thus,the roots are $2, 2, 3, 3$.
Let the roots be $r_1=2, r_2=2, r_3=3, r_4=3$.
We increase two distinct roots by $1$. The distinct roots are $2$ and $3$.
Increasing these by $1$ gives new roots $3$ and $4$.
The other two roots remain fixed as $2$ and $3$.
So the new roots are $2, 3, 3, 4$.
The original roots are ${2, 2, 3, 3}$ and the new roots are ${2, 3, 3, 4}$.
The common roots are $2, 3, 3$.
However,the question asks for the number of common roots.
Comparing the sets,the common values are $2$ and $3$.
Thus,there are $2$ distinct common roots.

(A) Given $f(x) = x^2 + bx + c$.
Since $f(1+k) = f(1-k)$ for all $k \in R$,the axis of symmetry of the parabola is $x = 1$.
The formula for the axis of symmetry of $f(x) = ax^2 + bx + c$ is $x = -b/(2a)$.
Here $a = 1$,so $-b/2 = 1$,which implies $b = -2$.
Thus,$f(x) = x^2 - 2x + c$.
Now,calculate the values:
$f(0) = 0^2 - 2(0) + c = c$.
$f(1) = 1^2 - 2(1) + c = 1 - 2 + c = c - 1$.
$f(-1) = (-1)^2 - 2(-1) + c = 1 + 2 + c = c + 3$.
Comparing these values: $c - 1 < c < c + 3$.
Therefore,$f(1) < f(0) < f(-1)$.

(D) Let $f(x) = 6x^3-25x^2+2x+8$. By the Rational Root Theorem,possible integer roots are factors of $8$ divided by factors of $6$. Testing $x=2$: $f(2) = 6(8)-25(4)+2(2)+8 = 48-100+4+8 = -40 \neq 0$. Testing $x=4$: $f(4) = 6(64)-25(16)+2(4)+8 = 384-400+8+8 = 0$. So,$x=4$ is a root. Dividing $6x^3-25x^2+2x+8$ by $(x-4)$,we get $6x^2-x-2=0$. Factoring this quadratic: $(2x+1)(3x-2)=0$. The roots are $x = -1/2$ and $x = 2/3$. Given $\alpha > 0$ and $\beta < 0$,we have $\alpha = 2/3$ and $\beta = -1/2$. Then,$\frac{4}{\alpha} + \frac{1}{\beta} = \frac{4}{2/3} + \frac{1}{-1/2} = 6 - 2 = 4$.

(D) Given equation is $x^4-x^2+x-1=0$.
Let $\alpha = \frac{1+\sqrt{3}i}{2}$. Since the coefficients are real,the conjugate $\beta = \frac{1-\sqrt{3}i}{2}$ must also be a root.
Sum of roots $\alpha + \beta = \frac{1+\sqrt{3}i + 1-\sqrt{3}i}{2} = 1$.
Product of roots $\alpha \beta = \frac{1^2 - (\sqrt{3}i)^2}{4} = \frac{1+3}{4} = 1$.
The quadratic factor corresponding to these roots is $x^2 - (\alpha+\beta)x + \alpha\beta = x^2 - x + 1 = 0$.
Dividing $x^4-x^2+x-1$ by $x^2-x+1$,we get $x^4-x^2+x-1 = (x^2-x+1)(x^2+x-1) = 0$.
The real roots are obtained from $x^2+x-1=0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.

(D) Since the coefficients $a$ and $b$ are real,the complex roots of the quadratic equation $x^2+ax+b=0$ must occur in conjugate pairs.
Given that $1-i$ is a root,its conjugate $1+i$ must also be a root of the equation.
For a quadratic equation $x^2+ax+b=0$,the product of the roots is given by the constant term $b$.
Therefore,$b = (1-i)(1+i)$.
Using the identity $(x-y)(x+y) = x^2-y^2$:
$b = 1^2 - i^2$.
Since $i^2 = -1$,we have:
$b = 1 - (-1) = 1 + 1 = 2$.

(A) Given $\sinh(\log x) = -2$.
Let $\log x = y$,then $x = e^y$.
Using the definition $\sinh y = \frac{e^y - e^{-y}}{2} = -2$.
$e^y - e^{-y} = -4$.
Multiply by $e^y$: $(e^y)^2 - 1 = -4e^y$.
$(e^y)^2 + 4e^y - 1 = 0$.
Using the quadratic formula $e^y = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$.
Since $x = e^y$ must be positive,$e^y = \sqrt{5} - 2$.
Thus,$x = \sqrt{5} - 2$.

(D) Given the equation $x^2 + 6x + 12 + 3 \sin(a + b\alpha) = 0$.
We can rewrite this as $x^2 + 6x + 9 + 3 + 3 \sin(a + b\alpha) = 0$.
$(x + 3)^2 + 3(1 + \sin(a + b\alpha)) = 0$.
Since $(x + 3)^2 \ge 0$ and $1 + \sin(a + b\alpha) \ge 0$ (because $-1 \le \sin \theta \le 1$),the sum of two non-negative terms can be zero only if each term is zero.
Therefore,$(x + 3)^2 = 0 \implies x = -3$ and $1 + \sin(a + b\alpha) = 0$.
This implies $\sin(a + b\alpha) = -1$.
For the least positive value of $\theta = a + b\alpha$,we have $\sin \theta = -1$,which gives $\theta = \frac{3\pi}{2}$.
We need to find $\cos(a + b\alpha) = \cos(\frac{3\pi}{2}) = 0$.

(A) Let $h(x) = f(x) - g(x) = (A - L)x^2 + (B - M)x - N$.
Given:
$h(2) = 4(A - L) + 2(B - M) - N = 1$ ... $(i)$
$h(3) = 9(A - L) + 3(B - M) - N = 4$ ... $(ii)$
$h(4) = 16(A - L) + 4(B - M) - N = 9$ ... $(iii)$
Subtracting $(i)$ from $(ii)$:
$5(A - L) + (B - M) = 3$ ... $(iv)$
Subtracting $(ii)$ from $(iii)$:
$7(A - L) + (B - M) = 5$ ... $(v)$
Subtracting $(iv)$ from $(v)$:
$2(A - L) = 2 \Rightarrow A - L = 1$.
Substituting $A - L = 1$ in $(iv)$:
$5(1) + (B - M) = 3 \Rightarrow B - M = -2$.
Substituting $A - L = 1$ and $B - M = -2$ in $(i)$:
$4(1) + 2(-2) - N = 1$ $\Rightarrow 4 - 4 - N = 1$ $\Rightarrow N = -1$.
Thus,$h(x) = (1)x^2 + (-2)x - (-1) = x^2 - 2x + 1$.
Setting $h(x) = 0$:
$x^2 - 2x + 1 = 0$ $\Rightarrow (x - 1)^2 = 0$ $\Rightarrow x = 1$.

(C) Given the equation $x^4-8x^3+11x^2+32x-60=0$.
By testing integer roots using the factor theorem,we find that $x = -2, 2, 3, 5$ are the roots.
Factoring the polynomial,we get $(x+2)(x-2)(x-3)(x-5)=0$.
Since $\alpha < \beta < \gamma < \delta$,we have $\alpha = -2, \beta = 2, \gamma = 3, \delta = 5$.
Now,calculating the expression $4\alpha+3\beta+2\gamma+\delta$:
$4(-2) + 3(2) + 2(3) + 5 = -8 + 6 + 6 + 5 = 9$.

(B) Given $\alpha$ and $\beta$ are roots of $x^2+ax-b=0$.
From Vieta's formulas,$\alpha+\beta = -a$ and $\alpha\beta = -b$.
Differentiating the curve $(\frac{x}{\alpha})^n + (\frac{y}{\beta})^n = 2$ with respect to $x$:
$\frac{n}{\alpha}(\frac{x}{\alpha})^{n-1} + \frac{n}{\beta}(\frac{y}{\beta})^{n-1} \frac{dy}{dx} = 0$.
At the point $(\alpha, \beta)$,the slope of the tangent is $\frac{dy}{dx} = -\frac{\beta}{\alpha}$.
The line $x \cos \theta + y \sin \theta = c$ has a slope of $-\cot \theta$.
Equating the slopes: $-\cot \theta = -\frac{\beta}{\alpha} \implies \cot \theta = \frac{\beta}{\alpha}$.
Since the line passes through $(\alpha, \beta)$,$\alpha \cos \theta + \beta \sin \theta = c$.
Using $\cot \theta = \frac{\beta}{\alpha}$,we have $\cos \theta = \frac{\beta}{\sqrt{\alpha^2+\beta^2}}$ and $\sin \theta = \frac{\alpha}{\sqrt{\alpha^2+\beta^2}}$.
Substituting these into the line equation: $\alpha(\frac{\beta}{\sqrt{\alpha^2+\beta^2}}) + \beta(\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}) = c \implies \frac{2\alpha\beta}{\sqrt{\alpha^2+\beta^2}} = c$.
We need to evaluate $(\frac{a}{b})^2 + \frac{2}{b} = (\frac{-(\alpha+\beta)}{-\alpha\beta})^2 + \frac{2}{-\alpha\beta} = \frac{(\alpha+\beta)^2}{(\alpha\beta)^2} - \frac{2}{\alpha\beta} = \frac{\alpha^2+\beta^2+2\alpha\beta-2\alpha\beta}{(\alpha\beta)^2} = \frac{\alpha^2+\beta^2}{(\alpha\beta)^2}$.
From $\frac{2\alpha\beta}{\sqrt{\alpha^2+\beta^2}} = c$,we get $\alpha^2+\beta^2 = \frac{4\alpha^2\beta^2}{c^2}$.
Substituting this into the expression: $\frac{4\alpha^2\beta^2/c^2}{(\alpha\beta)^2} = \frac{4}{c^2}$.

(B) For $n^3+\alpha n$ to be divisible by $3$ for all $n \in N$,we check $n=1$: $1^3+\alpha(1) = 1+\alpha$. For this to be divisible by $3$,the least positive integer $\alpha$ is $2$ (since $1+2=3$).
For $n^3-\beta n$ to be divisible by $6$ for all $n \in N$,we check $n=2$: $2^3-\beta(2) = 8-2\beta$. For this to be divisible by $6$,$8-2\beta = 6k$. If $\beta=1$,$8-2=6$,which is divisible by $6$. Thus,the least positive integer $\beta$ is $1$.
Therefore,$\alpha+\beta = 2+1 = 3$.

(C) The given equation is $||2x-3|-4|=2$.
This implies two cases:
Case $1$: $|2x-3|-4 = 2 \implies |2x-3| = 6$.
This further splits into:
$2x-3 = 6 \implies 2x = 9 \implies x = 4.5$.
$2x-3 = -6 \implies 2x = -3 \implies x = -1.5$.
Case $2$: $|2x-3|-4 = -2 \implies |2x-3| = 2$.
This further splits into:
$2x-3 = 2 \implies 2x = 5 \implies x = 2.5$.
$2x-3 = -2 \implies 2x = 1 \implies x = 0.5$.
The roots are $4.5, -1.5, 2.5, 0.5$.
The sum of the roots is $4.5 - 1.5 + 2.5 + 0.5 = 6$.

(A) Let the remainder be $ax+b$ when $P(x)$ is divided by $(x-3)(x-5)$.
$P(x) = (x-3)(x-5)Q(x) + (ax+b)$
Given that $P(3) = 10$ and $P(5) = 6$.
Substituting $x=3$ in the equation: $3a+b = 10$ $(i)$
Substituting $x=5$ in the equation: $5a+b = 6$ $(ii)$
Subtracting equation $(i)$ from $(ii)$:
$(5a+b) - (3a+b) = 6 - 10$
$2a = -4 \Rightarrow a = -2$
Substituting $a = -2$ in equation $(i)$:
$3(-2) + b = 10$
$-6 + b = 10 \Rightarrow b = 16$
Thus,the remainder is $-2x+16$.

(C) Since $a, b$ and $c$ are in $GP$,we have $b^{2} = ac$. Taking the natural logarithm on both sides,we get $2 \log_{e} b = \log_{e} a + \log_{e} c$.
Given the quadratic equation $(\log_{e} a) x^{2} - (2 \log_{e} b) x + \log_{e} c = 0$.
Substituting $x = 1$ into the equation,we get $(\log_{e} a) - (2 \log_{e} b) + \log_{e} c = 0$,which simplifies to $\log_{e} a + \log_{e} c = 2 \log_{e} b$,which is true.
Thus,$x = 1$ is one root of the equation.
Let the other root be $\alpha$. The product of the roots is given by $\frac{\log_{e} c}{\log_{e} a}$.
Therefore,$1 \times \alpha = \frac{\log_{e} c}{\log_{e} a} = \log_{a} c$.
Hence,the roots are $1$ and $\log_{a} c$.

(A) For the equation $ax^2 + bx + c = 0$ to have rational roots,the discriminant $D = b^2 - 4ac$ must be a perfect square of a rational number. Since $a, b, c$ are integers,$D$ must be a perfect square of an integer.
Given $a, b, c$ are odd natural numbers,$b^2$ is odd and $4ac$ is even. Thus,$D = b^2 - 4ac$ is an odd integer.
Let $D = (2k + 1)^2$ for some integer $k$. Then $b^2 - 4ac = (2k + 1)^2$.
Rearranging gives $4ac = b^2 - (2k + 1)^2 = (b - (2k + 1))(b + (2k + 1))$.
Since $b$ is odd,let $b = 2n + 1$. Then $4ac = (2n + 1 - 2k - 1)(2n + 1 + 2k + 1) = (2n - 2k)(2n + 2k + 2) = 4(n - k)(n + k + 1)$.
So,$ac = (n - k)(n + k + 1)$.
Note that $(n - k)$ and $(n + k + 1)$ differ by $(n + k + 1) - (n - k) = 2k + 1$,which is odd. Thus,one of $(n - k)$ or $(n + k + 1)$ must be even,making their product $ac$ even.
However,the product of two odd numbers $a$ and $c$ must be odd. This is a contradiction.
Therefore,$D$ cannot be a perfect square,and the equation has no rational roots.
Thus,the number of rational roots is $0$.

(B) The equation $P(x) \cdot Q(x) = 0$ implies that either $P(x) = 0$ or $Q(x) = 0$.
Let $D_1$ be the discriminant of $P(x) = ax^2 + bx + c$,so $D_1 = b^2 - 4ac$.
Let $D_2$ be the discriminant of $Q(x) = -ax^2 + dx + c$,so $D_2 = d^2 - 4(-a)(c) = d^2 + 4ac$.
Adding the two discriminants,we get $D_1 + D_2 = (b^2 - 4ac) + (d^2 + 4ac) = b^2 + d^2$.
Since $b^2 + d^2 \geq 0$,at least one of $D_1$ or $D_2$ must be non-negative.
If $D_1 \geq 0$,$P(x) = 0$ has at least two real roots (or one repeated root). If $D_2 \geq 0$,$Q(x) = 0$ has at least two real roots.
Since $ac \neq 0$,the quadratic terms exist. Thus,the equation $P(x) \cdot Q(x) = 0$ has at least two real roots.

(A) The given quadratic equation is $2ax^2 + (2a + b)x + b = 0$.
The discriminant $D$ is given by $D = B^2 - 4AC$.
Here,$A = 2a$,$B = (2a + b)$,and $C = b$.
$D = (2a + b)^2 - 4(2a)(b) = 4a^2 + 4ab + b^2 - 8ab = 4a^2 - 4ab + b^2 = (2a - b)^2$.
Since $a$ and $b$ are integers,$(2a - b)^2$ is a perfect square.
If the discriminant of a quadratic equation with rational coefficients is a perfect square,the roots are rational.
Thus,the roots are rational.

(A) Given $P(x) = ax^{2} + bx + c$ and $Q(x) = -ax^{2} + dx + c$.
For $P(x) = 0$,the discriminant is $D_{1} = b^{2} - 4ac$.
For $Q(x) = 0$,the discriminant is $D_{2} = d^{2} - 4(-a)(c) = d^{2} + 4ac$.
Adding the two discriminants,we get $D_{1} + D_{2} = b^{2} + d^{2} \geq 0$.
Since the sum of the discriminants is non-negative,at least one of the discriminants must be non-negative ($D_{1} \geq 0$ or $D_{2} \geq 0$).
If $D_{1} \geq 0$,$P(x)$ has real roots. If $D_{2} \geq 0$,$Q(x)$ has real roots.
Therefore,the equation $P(x) \cdot Q(x) = 0$ must have at least two real roots.

(A) Consider the two quadratic equations:
$x^{2} + b_{1}x + c_{1} = 0$ and $x^{2} + b_{2}x + c_{2} = 0$.
Let $D_{1}$ and $D_{2}$ be the discriminants of these equations respectively.
$D_{1} = b_{1}^{2} - 4c_{1}$
$D_{2} = b_{2}^{2} - 4c_{2}$
Adding the two discriminants,we get:
$D_{1} + D_{2} = b_{1}^{2} + b_{2}^{2} - 4(c_{1} + c_{2})$
Given that $b_{1}b_{2} = 2(c_{1} + c_{2})$,we substitute $4(c_{1} + c_{2}) = 2b_{1}b_{2}$ into the equation:
$D_{1} + D_{2} = b_{1}^{2} + b_{2}^{2} - 2b_{1}b_{2}$
$D_{1} + D_{2} = (b_{1} - b_{2})^{2}$
Since $(b_{1} - b_{2})^{2} \geq 0$ for all real $b_{1}, b_{2}$,it follows that $D_{1} + D_{2} \geq 0$.
If the sum of two real numbers is non-negative,then at least one of them must be non-negative.
Therefore,at least one of $D_{1}$ or $D_{2}$ is $\geq 0$,which implies that at least one of the equations has real roots.

(A) The given quadratic equation is $2px^{2} + (2p + q)x + q = 0$.
The discriminant $D$ is given by $D = b^{2} - 4ac$.
Substituting the coefficients $a = 2p$,$b = (2p + q)$,and $c = q$:
$D = (2p + q)^{2} - 4(2p)(q)$
$D = 4p^{2} + q^{2} + 4pq - 8pq$
$D = 4p^{2} + q^{2} - 4pq$
$D = (2p - q)^{2}$
Since $p$ and $q$ are integers,$D$ is a perfect square of an integer.
For a quadratic equation with rational coefficients,if the discriminant is a perfect square,the roots are rational.

(C) The given quadratic equation is $ax^{2} + bx + 1 = 0$.
For the equation to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = b^{2} - 4ac \geq 0$.
Substituting $c = 1$,we get $b^{2} - 4a \geq 0$,which implies $b^{2} \geq 4a$.
Given $a, b \in \{1, 2, 3\}$,we test the possible values:
If $a = 1$,$b^{2} \geq 4 \implies b \in \{2, 3\}$. Pairs: $(1, 2), (1, 3)$.
If $a = 2$,$b^{2} \geq 8 \implies b = 3$. Pair: $(2, 3)$.
If $a = 3$,$b^{2} \geq 12 \implies$ no value of $b$ satisfies this.
The possible ordered pairs $(a, b)$ are $(1, 2), (1, 3), (2, 3)$.
Thus,the number of possible ordered pairs is $3$.

(C) For a quadratic equation $ax^2 + bx + c = 0$ with irrational coefficients,the roots are not necessarily irrational in pairs.
For example,consider the equation $x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0$.
The roots are $x = 1$ and $x = \sqrt{2}$.
Here,one root is rational and one is irrational.
Thus,the statement that such an equation must have zero or two irrational roots is false.
Therefore,option $C$ is always false.

(A) Given,$\alpha, \beta$ are the roots of $ax^{2}+bx+c=0$ and $\alpha+h, \beta+h$ are the roots of $px^{2}+qx+r=0$.
For the first equation,the discriminant is $D_{1} = b^{2}-4ac$ and the difference of roots is $(\alpha-\beta)^{2} = \frac{D_{1}}{a^{2}}$.
For the second equation,the discriminant is $D_{2} = q^{2}-4pr$ and the difference of roots is $((\alpha+h)-(\beta+h))^{2} = \frac{D_{2}}{p^{2}}$.
Since $(\alpha+h)-(\beta+h) = \alpha-\beta$,we have $(\alpha-\beta)^{2} = ((\alpha+h)-(\beta+h))^{2}$.
Therefore,$\frac{D_{1}}{a^{2}} = \frac{D_{2}}{p^{2}}$.
This implies $\frac{D_{1}}{D_{2}} = \frac{a^{2}}{p^{2}}$.
Thus,the ratio of the discriminants is $a^{2}: p^{2}$.

(A) Since $a, b$ and $c$ are in $AP$,we have $2b = a + c$.
Given the quadratic equation $a x^{2} - 2b x + c = 0$.
Substituting $2b = a + c$ into the equation,we get:
$a x^{2} - (a + c) x + c = 0$
$a x^{2} - a x - c x + c = 0$
$a x(x - 1) - c(x - 1) = 0$
$(x - 1)(a x - c) = 0$
Thus,the roots are $x = 1$ and $x = \frac{c}{a}$.

(C) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
Here,$a = 1$,$b = -2\sqrt{3}$,and $c = -22$.
$D = (-2\sqrt{3})^2 - 4(1)(-22) = 12 + 88 = 100$.
Since $D > 0$,the roots are real and unequal.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{2\sqrt{3} \pm \sqrt{100}}{2} = \frac{2\sqrt{3} \pm 10}{2} = \sqrt{3} \pm 5$.
Since $\sqrt{3}$ is an irrational number,the roots $\sqrt{3} + 5$ and $\sqrt{3} - 5$ are irrational.
Therefore,the roots are real,irrational,and unequal.

(B) Let $|x-2| = y$. Since $|x-2| \ge 0$,we must have $y \ge 0$.
Substituting into the equation,we get $y^2 + y - 2 = 0$.
Factoring the quadratic equation: $(y+2)(y-1) = 0$.
This gives $y = -2$ or $y = 1$.
Since $y \ge 0$,we discard $y = -2$. Thus,$y = 1$.
Now,solve $|x-2| = 1$:
$x-2 = 1$ or $x-2 = -1$.
$x = 3$ or $x = 1$.
The real roots are $3$ and $1$.
The sum of the roots is $3 + 1 = 4$.

(A) Let the roots of the given equation $3ax^2+3bx+c=0$ be $\alpha$ and $\beta$.
Then,$3a\alpha^2+3b\alpha+c=0$.
We want to find the equation whose roots are $3\alpha$ and $3\beta$. Let $x = 3\alpha$,which implies $\alpha = \frac{x}{3}$.
Substituting $\alpha = \frac{x}{3}$ into the original equation:
$3a(\frac{x}{3})^2 + 3b(\frac{x}{3}) + c = 0$
$3a(\frac{x^2}{9}) + bx + c = 0$
$\frac{ax^2}{3} + bx + c = 0$
Multiplying the entire equation by $3$,we get:
$ax^2 + 3bx + 3c = 0$.

(C) Expanding the given equation:
$(x^2 - (b+c)x + bc) + (x^2 - (a+c)x + ac) + (x^2 - (a+b)x + ab) = 0$
$3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0$
The discriminant $D$ of this quadratic equation is given by:
$D = [-2(a+b+c)]^2 - 4(3)(ab+bc+ca)$
$D = 4(a+b+c)^2 - 12(ab+bc+ca)$
$D = 4(a^2+b^2+c^2 + 2ab+2bc+2ca - 3ab-3bc-3ca)$
$D = 4(a^2+b^2+c^2 - ab-bc-ca)$
$D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]$
Since $a, b, c$ are real,$(a-b)^2, (b-c)^2, (c-a)^2 \geq 0$.
Therefore,$D \geq 0$.
Since the discriminant is always non-negative,the roots are always real.

(C) For a quadratic expression $f(x) = ax^{2} + bx + c$ to have the same sign as $a$ for all $x$,the parabola must not intersect the $x$-axis.
This implies that the quadratic equation $ax^{2} + bx + c = 0$ has no real roots.
For no real roots,the discriminant $D = b^{2} - 4ac$ must be less than $0$.
Therefore,the condition is $b^{2} - 4ac < 0$.

(C) Given the equation $x^2+15|x|+14=0$.
Since $x^2 = |x|^2$,we can rewrite the equation as $|x|^2+15|x|+14=0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2+15t+14=0$.
Factoring the quadratic,we get $(t+1)(t+14)=0$.
This gives $t = -1$ or $t = -14$.
However,we defined $t = |x|$,which must be non-negative $(t \ge 0)$.
Since both $-1$ and $-14$ are less than $0$,there is no real value of $x$ that satisfies the equation.
Thus,the equation has no solution.

(B) For a quadratic equation $Ax^{2} + Bx + C = 0$ to have roots of opposite signs,the product of the roots must be less than zero.
Product of the roots = $\frac{C}{A} < 0$.
Here,$A = 2$ and $C = a^{2} - 4a$.
So,$\frac{a^{2} - 4a}{2} < 0$.
$a^{2} - 4a < 0$.
$a(a - 4) < 0$.
This inequality holds when $0 < a < 4$.

(B) Given equation is $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$.
For the square roots to be defined,we must have $x+1 \ge 0$,$x-1 \ge 0$,and $4x-1 \ge 0$,which implies $x \ge 1$.
Squaring both sides,we get $(x+1) + (x-1) - 2\sqrt{x^2-1} = 4x-1$.
This simplifies to $2x - 2\sqrt{x^2-1} = 4x-1$,or $-2\sqrt{x^2-1} = 2x-1$.
Squaring both sides again,we get $4(x^2-1) = (2x-1)^2 = 4x^2 - 4x + 1$.
This simplifies to $4x^2 - 4 = 4x^2 - 4x + 1$,which gives $4x = 5$,or $x = \frac{5}{4}$.
Checking $x = \frac{5}{4}$ in the original equation: $\sqrt{\frac{5}{4}+1} - \sqrt{\frac{5}{4}-1} = \sqrt{\frac{9}{4}} - \sqrt{\frac{1}{4}} = \frac{3}{2} - \frac{1}{2} = 1$.
However,the right side is $\sqrt{4(\frac{5}{4})-1} = \sqrt{5-1} = \sqrt{4} = 2$.
Since $1 \neq 2$,$x = \frac{5}{4}$ is an extraneous solution.
Therefore,there are no real solutions to the equation.

(C) quadratic equation is given by $ax^2 + bx + c = 0$,where $a \neq 0$.
Since $0$ is a solution,substituting $x = 0$ into the equation gives $a(0)^2 + b(0) + c = 0$,which implies $c = 0$.
The coefficients $a, b, c$ are chosen from the set $\{0, 1, 2, \dots, 9\}$.
For the equation to be quadratic,the coefficient $a$ must be non-zero,so $a \in \{1, 2, \dots, 9\}$. This gives $9$ possible choices for $a$.
The coefficient $b$ can be any value from the set $\{0, 1, 2, \dots, 9\}$,which gives $10$ possible choices for $b$.
The coefficient $c$ is fixed as $0$,so there is only $1$ choice for $c$.
Therefore,the total number of such quadratic equations is $N = 9 \times 10 \times 1 = 90$.

(C) For the quadratic equation $ax^2 + 2bx + c = 0$ to have no real roots,the discriminant $D < 0$.
$D = (2b)^2 - 4ac < 0 \implies 4b^2 < 4ac \implies b^2 < ac$.
Since $a, b, c > 0$,we have $b < \sqrt{ac}$.
$1$. If $a, b, c$ are in $A$.$P$.,then $2b = a + c$. Since $b < \sqrt{ac}$,we have $a + c < 2\sqrt{ac}$,which implies $(\sqrt{a} - \sqrt{c})^2 < 0$,which is impossible. Thus,$a, b, c$ cannot be in $A$.$P$.
$2$. If $a, b, c$ are in $G$.$P$.,then $b^2 = ac$. But we have $b^2 < ac$,so $a, b, c$ cannot be in $G$.$P$.
$3$. If $a, b, c$ are in $H$.$P$.,then $b = \frac{2ac}{a+c}$. Since $a, c > 0$,by $A$.$M$. $\geq$ $G$.$M$.,we have $\frac{a+c}{2} \geq \sqrt{ac}$,so $b = \frac{2ac}{a+c} \leq \sqrt{ac}$. The condition $b < \sqrt{ac}$ is satisfied if $a \neq c$. Thus,$a, b, c$ can be in $H$.$P$.

(B) Step $1$. Rewrite the equation using the identity $\cos(2x) = 1 - 2\sin^2 x$:
$1 - 2\sin^2 x = 2a - a\sin x$
$2\sin^2 x - a\sin x + 2a - 1 = 0$
Wait,let us re-evaluate the given equation $\cos(2x + 7) = a(2 - \sin x)$.
Actually,the standard form is $2\sin^2 x - a\sin x + (2a - 8) = 0$.
Step $2$. Solve for $\sin x$ using the quadratic formula:
$\sin x = \frac{a \pm \sqrt{a^2 - 4(2)(2a - 8)}}{2(2)} = \frac{a \pm \sqrt{a^2 - 16a + 64}}{4} = \frac{a \pm \sqrt{(a - 8)^2}}{4} = \frac{a \pm (a - 8)}{4}$.
This gives two roots:
$\sin x = \frac{a + a - 8}{4} = \frac{2a - 8}{4} = \frac{a - 4}{2}$
$\sin x = \frac{a - a + 8}{4} = \frac{8}{4} = 2$ (Not possible as $\sin x \leq 1$).
Step $3$. Since $-1 \leq \sin x \leq 1$,we have:
$-1 \leq \frac{a - 4}{2} \leq 1$
$-2 \leq a - 4 \leq 2$
$2 \leq a \leq 6$.
Thus,$a \in [2, 6]$.

(C) Given the equation $ax^2 + bx + c = 0$ and the condition $a + 2b + 4c = 0$.
Divide the condition $a + 2b + 4c = 0$ by $4$:
$\frac{a}{4} + \frac{2b}{4} + \frac{4c}{4} = 0$
$\frac{a}{4} + \frac{b}{2} + c = 0$
This can be rewritten as:
$a(\frac{1}{2})^2 + b(\frac{1}{2}) + c = 0$
Comparing this with the quadratic equation $f(x) = ax^2 + bx + c = 0$,we see that $f(\frac{1}{2}) = 0$.
Therefore,$x = \frac{1}{2}$ is a root of the equation.

(C) Let $h(x) = f(x) - g(x) = (a-p)x^2 + (b-q)x + (c-r)$.
Since $f(1) = g(1)$,we have $h(1) = 0$.
Since $f(2) = g(2)$,we have $h(2) = 0$.
Since $h(x)$ is a quadratic polynomial with roots $1$ and $2$,we can write $h(x) = k(x-1)(x-2)$ for some constant $k$.
We are given $f(3) - g(3) = 2$,which means $h(3) = 2$.
Substituting $x=3$ into $h(x) = k(x-1)(x-2)$,we get $h(3) = k(3-1)(3-2) = k(2)(1) = 2k$.
Since $h(3) = 2$,we have $2k = 2$,which implies $k = 1$.
Thus,$h(x) = 1(x-1)(x-2) = (x-1)(x-2)$.
We need to find $f(4) - g(4)$,which is $h(4)$.
$h(4) = (4-1)(4-2) = (3)(2) = 6$.

(D) Let $P(x) = x^{3} + ax^{2} + bx + c$. For $P(x)$ to be divisible by $x^{2} + 2$,we perform polynomial division or equate coefficients.
Dividing $x^{3} + ax^{2} + bx + c$ by $x^{2} + 2$ gives a quotient of $(x + a)$ and a remainder of $(b - 2)x + (c - 2a)$.
For the polynomial to be divisible,the remainder must be zero,so $(b - 2)x + (c - 2a) = 0$.
This implies $b - 2 = 0$ and $c - 2a = 0$.
Thus,$b = 2$ and $c = 2a$.
Given $a, b, c \in \mathbb{N}$ and $a, b, c \le 20$,we have $b = 2$ (which is fixed).
For $c = 2a$,since $c \le 20$,we have $2a \le 20$,which means $a \le 10$.
Since $a \in \mathbb{N}$,$a$ can take values from $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Therefore,there are $10$ such polynomials.

(A) To find the number of real solutions for $x|x+3|+|x-1|-2=0$,we analyze the equation in three intervals based on the critical points $x=-3$ and $x=1$:
$I$. Case $x \ge 1$:
$x(x+3) + (x-1) - 2 = 0 \implies x^2+3x+x-1-2=0 \implies x^2+4x-3=0$.
Using the quadratic formula,$x = \frac{-4 \pm \sqrt{16 - 4(1)(-3)}}{2} = -2 \pm \sqrt{7}$.
Since $x \ge 1$,both $-2+\sqrt{7} \approx 0.646$ and $-2-\sqrt{7} \approx -4.646$ are rejected.
$II$. Case $-3 \le x < 1$:
$x(x+3) - (x-1) - 2 = 0 \implies x^2+3x-x+1-2=0 \implies x^2+2x-1=0$.
Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = -1 \pm \sqrt{2}$.
Since $-3 \le x < 1$,both $x = -1+\sqrt{2} \approx 0.414$ and $x = -1-\sqrt{2} \approx -2.414$ are valid solutions.
$III$. Case $x < -3$:
$x(-(x+3)) - (x-1) - 2 = 0 \implies -x^2-3x-x+1-2=0 \implies -x^2-4x-1=0 \implies x^2+4x+1=0$.
Using the quadratic formula,$x = \frac{-4 \pm \sqrt{16 - 4(1)(1)}}{2} = -2 \pm \sqrt{3}$.
Since $x < -3$,both $-2+\sqrt{3} \approx -0.268$ and $-2-\sqrt{3} \approx -3.732$ are considered. Only $x = -2-\sqrt{3}$ is valid as it is less than $-3$.
Thus,the valid solutions are $x = -1+\sqrt{2}$,$x = -1-\sqrt{2}$,and $x = -2-\sqrt{3}$. The total number of real solutions is $3$.

(D) $x^4 - ax^2 + 9 = 0$ . . . . $(1)$
Let $x^2 = t$.
Then $t^2 - at + 9 = 0$ . . . . $(2)$
For the roots of equation $(1)$ to be real and distinct,the roots of equation $(2)$ must be positive and distinct.
$(i)$ Discriminant $D > 0$ $\Rightarrow a^2 - 36 > 0$ $\Rightarrow a \in (-\infty, -6) \cup (6, \infty)$.
$(ii)$ Sum of roots $\frac{-b}{a} > 0 \Rightarrow a > 0$.
$(iii)$ Product of roots $\frac{c}{a} > 0 \Rightarrow 9 > 0$,which is true for all $a \in \mathbb{R}$.
By taking the intersection of $(i), (ii),$ and $(iii)$,we get $a > 6$.
Therefore,the smallest positive integral value of $a$ is $7$.

(A) The given equation is $\sum_{r=1}^{n}(x+2r-2)(x+2r) = \frac{8n}{3}$.
Expanding the terms: $\sum_{r=1}^{n}(x^2 + 4rx - 2x + 4r^2 - 4r) = \frac{8n}{3}$.
This simplifies to $nx^2 + 2x(2\sum r - n) + 4\sum r^2 - 4\sum r = \frac{8n}{3}$.
Using $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$,we get $nx^2 + 2x(n^2) + \frac{4n(n+1)(2n+1)}{6} - 2n(n+1) = \frac{8n}{3}$.
Dividing by $n$: $x^2 + 2nx + \frac{2(n+1)(2n+1)}{3} - 2(n+1) = \frac{8}{3}$.
$x^2 + 2nx + \frac{4n^2+6n+2-6n-6-8}{3} = 0 \Rightarrow x^2 + 2nx + \frac{4n^2-12}{3} = 0$.
Since the roots $\alpha, \beta$ are consecutive even integers,$|\alpha - \beta| = 2$.
Thus,$\frac{\sqrt{D}}{1} = 2 \Rightarrow D = 4$.
$D = (2n)^2 - 4(\frac{4n^2-12}{3}) = 4$.
$4n^2 - \frac{16n^2-48}{3} = 4 \Rightarrow 12n^2 - 16n^2 + 48 = 12$.
$-4n^2 = -36$ $\Rightarrow n^2 = 9$ $\Rightarrow n = 3$.

(C) Given the quadratic equation $12x^{2}-20x+3\lambda=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha+\beta = \frac{20}{12} = \frac{5}{3}$ and $\alpha\beta = \frac{3\lambda}{12} = \frac{\lambda}{4}$.
We are given $\frac{1}{2} \le |\beta-\alpha| \le \frac{3}{2}$.
Squaring the inequality,we get $\frac{1}{4} \le (\beta-\alpha)^{2} \le \frac{9}{4}$.
Using the identity $(\beta-\alpha)^{2} = (\alpha+\beta)^{2} - 4\alpha\beta$,we have:
$\frac{1}{4} \le (\frac{5}{3})^{2} - 4(\frac{\lambda}{4}) \le \frac{9}{4}$.
$\frac{1}{4} \le \frac{25}{9} - \lambda \le \frac{9}{4}$.
Subtracting $\frac{25}{9}$ from all parts:
$\frac{1}{4} - \frac{25}{9} \le -\lambda \le \frac{9}{4} - \frac{25}{9}$.
$\frac{9-100}{36} \le -\lambda \le \frac{81-100}{36}$.
$-\frac{91}{36} \le -\lambda \le -\frac{19}{36}$.
Multiplying by $-1$ reverses the inequality:
$\frac{19}{36} \le \lambda \le \frac{91}{36}$.
Since $\frac{19}{36} \approx 0.527$ and $\frac{91}{36} \approx 2.527$,the possible integer values for $\lambda$ are $1$ and $2$.
The sum of these values is $1+2 = 3$.

(D) Let the time taken by mason $A$ alone to complete the work be $x$ days. Then,mason $B$ alone takes $x+24$ days.
Work done by $A$ in $1$ day $= \frac{1}{x}$.
Work done by $B$ in $1$ day $= \frac{1}{x+24}$.
Work done by $A+B$ in $1$ day $= \frac{1}{22.5} = \frac{1}{45/2} = \frac{2}{45}$.
So,$\frac{1}{x} + \frac{1}{x+24} = \frac{2}{45}$.
$\frac{x+24+x}{x(x+24)} = \frac{2}{45} \implies \frac{2x+24}{x^2+24x} = \frac{2}{45}$.
$45(x+12) = x^2+24x \implies 45x + 540 = x^2 + 24x$.
$x^2 - 21x - 540 = 0$.
$(x-36)(x+15) = 0$.
Since time cannot be negative,$x = 36$ days.

(C) The given equation is $(\sqrt{3}-2)|\sqrt{x}-3| + (\sqrt{x}-3)^2 - 2\sqrt{3} = 0$.
Let $t = |\sqrt{x}-3|$. Then the equation becomes $t^2 + (\sqrt{3}-2)t - 2\sqrt{3} = 0$.
Factoring the quadratic: $(t+ \sqrt{3})(t-2) = 0$.
Since $t = |\sqrt{x}-3| \ge 0$,we must have $t = 2$.
Thus,$|\sqrt{x}-3| = 2$,which implies $\sqrt{x}-3 = 2$ or $\sqrt{x}-3 = -2$.
This gives $\sqrt{x} = 5$ or $\sqrt{x} = 1$.
So,$x = 25$ or $x = 1$.
Given $\alpha < \beta$,we have $\alpha = 1$ and $\beta = 25$.
Finally,$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{\frac{25}{1}} + \sqrt{1 \times 25} = 5 + 5 = 10$.

(B) We analyze the equation $x|x+4|+3|x+2|+10=0$ by considering different intervals for $x$:
Case $I$: $x < -4$. The equation becomes $x(-(x+4)) + 3(-(x+2)) + 10 = 0$,which simplifies to $-x^2 - 4x - 3x - 6 + 10 = 0$,or $x^2 + 7x - 4 = 0$. The roots are $x = \frac{-7 \pm \sqrt{49 - 4(1)(-4)}}{2} = \frac{-7 \pm \sqrt{65}}{2}$. Since $\sqrt{65} \approx 8.06$,$x_1 = \frac{-7 - 8.06}{2} \approx -7.53$ (which is $< -4$,so it is a valid solution) and $x_2 = \frac{-7 + 8.06}{2} \approx 0.53$ (which is not $< -4$).
Case $II$: $-4 \leq x < -2$. The equation becomes $x(x+4) + 3(-(x+2)) + 10 = 0$,which simplifies to $x^2 + 4x - 3x - 6 + 10 = 0$,or $x^2 + x + 4 = 0$. The discriminant $D = 1^2 - 4(1)(4) = -15 < 0$,so there are no real solutions.
Case $III$: $x \geq -2$. The equation becomes $x(x+4) + 3(x+2) + 10 = 0$,which simplifies to $x^2 + 4x + 3x + 6 + 10 = 0$,or $x^2 + 7x + 16 = 0$. The discriminant $D = 7^2 - 4(1)(16) = 49 - 64 = -15 < 0$,so there are no real solutions.
Thus,there is only $1$ distinct real solution.

(A) Let $|x-1|=t$.
Then the equation becomes $t^{2}-5t+6=0$.
Factoring the quadratic equation,we get $(t-2)(t-3)=0$,so $t=2$ or $t=3$.
Case $1$: $|x-1|=2 \implies x-1=2$ or $x-1=-2$,which gives $x=3$ or $x=-1$.
Case $2$: $|x-1|=3 \implies x-1=3$ or $x-1=-3$,which gives $x=4$ or $x=-2$.
The roots are $3, -1, 4, -2$.
The sum of the roots is $3 + (-1) + 4 + (-2) = 4$.