Quadratic expressions and Position of roots Questions in English

(C) For the roots of the quadratic equation $x^2-(p+2)x+(2p+9)=0$ to be negative real numbers,the following conditions must be satisfied:
$1$. The discriminant $D \geq 0$:
$D = (p+2)^2 - 4(2p+9) \geq 0$
$p^2 + 4p + 4 - 8p - 36 \geq 0$
$p^2 - 4p - 32 \geq 0$
$(p-8)(p+4) \geq 0$
This implies $p \in (-\infty, -4] \cup [8, \infty)$.
$2$. The sum of roots must be negative:
Sum $= -\frac{b}{a} = p+2 < 0 \implies p < -2$.
$3$. The product of roots must be positive:
Product $= \frac{c}{a} = 2p+9 > 0 \implies p > -\frac{9}{2}$.
Combining these conditions: $p \in (-\frac{9}{2}, -4]$.
Thus,$\alpha = -\frac{9}{2}$ and $\beta = -4$.
We need to find $\beta - 2\alpha = -4 - 2(-\frac{9}{2}) = -4 + 9 = 5$.

(A) Let $f(x) = x^2 - 3x + k$. For the equation $f(x) = 0$ to have at least one root in the interval $[0, 1]$,we consider the following conditions:
$1$. The product of the values at the endpoints must be less than or equal to zero: $f(0) \cdot f(1) \le 0$.
$f(0) = 0^2 - 3(0) + k = k$.
$f(1) = 1^2 - 3(1) + k = k - 2$.
So,$k(k - 2) \le 0$,which implies $0 \le k \le 2$.
$2$. The vertex of the parabola $x = -b/(2a) = 3/2$ does not lie in the interval $[0, 1]$.
$3$. Since the parabola opens upward,if the vertex is outside the interval,the only way to have a root in $[0, 1]$ is if the function changes sign at the endpoints.
Therefore,the required range for $k$ is $0 \le k \le 2$.

(D) Given $ax^2 + bx + c < 0$ for all $x \in R$. This implies $a < 0$ and $D = b^2 - 4ac < 0$.
The extreme value of $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.
The extreme value of $cx^2 + ax + b$ occurs at $x = -\frac{a}{2c}$.
Since these points are the same,$-\frac{b}{2a} = -\frac{a}{2c}$,which implies $a^2 = bc$.
Since $a < 0$ and $a^2 = bc$,$c$ must also be negative.
The expression $cx^2 + ax + b$ has a maximum value because $c < 0$.
The maximum value is given by $-\frac{D'}{4c} = -\frac{a^2 - 4bc}{4c} = -\frac{bc - 4bc}{4c} = -\frac{-3bc}{4c} = \frac{3b}{4}$.
Thus,the maximum value is $\frac{3b}{4}$.

(B) Given $f(x) = x^2 + 2kx + k < 0$ for all $x \in [1, 2]$.
Since the coefficient of $x^2$ is positive,the parabola opens upward.
For a quadratic $f(x)$ to be negative on $[1, 2]$,the maximum value of $f(x)$ on the interval $[1, 2]$ must be less than $0$.
Since the parabola opens upward,the maximum value on $[1, 2]$ occurs at the endpoints $x=1$ or $x=2$.
$f(1) = 1^2 + 2k(1) + k = 1 + 3k < 0 \implies 3k < -1 \implies k < -1/3$.
$f(2) = 2^2 + 2k(2) + k = 4 + 5k < 0 \implies 5k < -4 \implies k < -4/5$.
For $f(x) < 0$ to hold for all $x \in [1, 2]$,both conditions must be satisfied simultaneously.
Thus,$k < \min(-1/3, -4/5)$,which gives $k < -4/5$.
Therefore,the interval for $k$ is $(-\infty, -4/5)$.

(C) The minimum value of $f(x) = x^2 + 2bx + 2c^2$ is found at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$.
The maximum value of $g(x) = -x^2 - 2cx + b^2$ is found at $x = -c$,which is $g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$.
Given $\min f(x) > \max g(x)$,we have $2c^2 - b^2 > c^2 + b^2$.
This simplifies to $c^2 > 2b^2$,or $\frac{c^2}{b^2} > 2$.
Taking the square root of both sides,we get $\left|\frac{c}{b}\right| > \sqrt{2}$.
Thus,$\left|\frac{c}{b}\right| \in (\sqrt{2}, \infty)$.

(D) The minimum value of a quadratic expression $ax^2 + bx + c$ (where $a > 0$) is given by the formula $\frac{4ac - b^2}{4a}$.
$i) x^2 + 4x + 6$: Here $a=1, b=4, c=6$. Minimum value = $\frac{4(1)(6) - (4)^2}{4(1)} = \frac{24 - 16}{4} = \frac{8}{4} = 2$. So,$i \rightarrow b$.
$ii) x^2 - 2x + 5$: Here $a=1, b=-2, c=5$. Minimum value = $\frac{4(1)(5) - (-2)^2}{4(1)} = \frac{20 - 4}{4} = \frac{16}{4} = 4$. So,$ii \rightarrow c$.
$iii) x^2 + 6x + 18$: Here $a=1, b=6, c=18$. Minimum value = $\frac{4(1)(18) - (6)^2}{4(1)} = \frac{72 - 36}{4} = \frac{36}{4} = 9$. So,$iii \rightarrow d$.
$iv) x^2 - 4x + 5$: Here $a=1, b=-4, c=5$. Minimum value = $\frac{4(1)(5) - (-4)^2}{4(1)} = \frac{20 - 16}{4} = \frac{4}{4} = 1$. So,$iv \rightarrow a$.
Thus,the correct matching is $i$ $\rightarrow b, ii$ $\rightarrow c, iii$ $\rightarrow d, iv$ $\rightarrow a$.

(A) The given quadratic expression is $f(x) = 2kx^2 - (4k+1)x + 2$.
We can factorize this as $f(x) = (2x - 1)(kx - 2)$.
The roots of $f(x) = 0$ are $x = \frac{1}{2}$ and $x = \frac{2}{k}$.
For the expression to be negative,$x$ must lie between the roots.
Case $1$: If $k > 0$,then $\frac{1}{2} < x < \frac{2}{k}$. The integers $x$ satisfying this are $1, 2, 3$.
For exactly three integers to exist,we need $3 < \frac{2}{k} \le 4$.
Solving $3 < \frac{2}{k}$ gives $k < \frac{2}{3}$.
Solving $\frac{2}{k} \le 4$ gives $k \ge \frac{1}{2}$.
So,$k \in [\frac{1}{2}, \frac{2}{3})$.
Case $2$: If $k < 0$,then $\frac{2}{k} < x < \frac{1}{2}$. The integers $x$ are $-1, -2, -3$.
For exactly three integers,we need $-4 \le \frac{2}{k} < -3$.
Solving $\frac{2}{k} < -3$ gives $k > -\frac{2}{3}$.
Solving $-4 \le \frac{2}{k}$ gives $k \le -\frac{1}{2}$.
So,$k \in [-\frac{2}{3}, -\frac{1}{2}]$.
Since the provided options do not match these intervals,the question or options may be flawed.

(D) Given equation: $x^2 - 6ax + (9a^2 - 2a + 2) = 0$.
Let $f(x) = x^2 - 6ax + 9a^2 - 2a + 2$.
For both roots to be greater than $3$,the following conditions must be satisfied:
$1)$ Discriminant $D \ge 0$:
$D = (-6a)^2 - 4(1)(9a^2 - 2a + 2) = 36a^2 - 36a^2 + 8a - 8 = 8a - 8$.
$8a - 8 \ge 0 \Rightarrow a \ge 1$.
$2)$ Vertex location: $-\frac{b}{2a} > 3$:
$-\frac{-6a}{2(1)} > 3$ $\Rightarrow 3a > 3$ $\Rightarrow a > 1$.
$3)$ $f(3) > 0$:
$f(3) = 3^2 - 6a(3) + 9a^2 - 2a + 2 > 0$
$9 - 18a + 9a^2 - 2a + 2 > 0$
$9a^2 - 20a + 11 > 0$
$(9a - 11)(a - 1) > 0$.
Since $a > 1$,the condition $(9a - 11)(a - 1) > 0$ implies $a > \frac{11}{9}$.
Combining all conditions,we get $a > \frac{11}{9}$.

(A) Let $f(x) = x^2+bx+c$ and $g(x) = x^2+b_1x+c_1$. The intersection point $P$ of the two parabolas is found by solving $f(x) = g(x)$.
$x^2+bx+c = x^2+b_1x+c_1$
$(b-b_1)x = c_1-c$
$x = \frac{c_1-c}{b-b_1} = \frac{c-c_1}{b_1-b}$.
Since $\gamma < \alpha < \delta < \beta$,the point of intersection $P$ lies below the $x$-axis,meaning $f(x) < 0$ at this $x$-coordinate.
$f\left(\frac{c-c_1}{b_1-b}\right) < 0$
$\left(\frac{c-c_1}{b_1-b}\right)^2 + b\left(\frac{c-c_1}{b_1-b}\right) + c < 0$
Multiplying by $(b_1-b)^2$ (which is positive):
$(c-c_1)^2 + b(c-c_1)(b_1-b) + c(b_1-b)^2 < 0$
$(c-c_1)^2 < -b(c-c_1)(b_1-b) - c(b_1-b)^2$
$(c-c_1)^2 < (b_1-b)[-b(c-c_1) - c(b_1-b)]$
$(c-c_1)^2 < (b_1-b)[-bc+bc_1-cb_1+cb]$
$(c-c_1)^2 < (b_1-b)(bc_1-b_1c)$.
Thus,option $A$ is correct.

(A) The function $f(x) = x^2 + 2bx + 2c^2$ is an upward-opening parabola. Its minimum value occurs at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$.
The function $g(x) = -x^2 - 2cx + b^2$ is a downward-opening parabola. Its maximum value occurs at $x = -c$,which is $g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$.
According to the problem,the minimum value of $f(x)$ is greater than the maximum value of $g(x)$:
$2c^2 - b^2 > c^2 + b^2$
$2c^2 - c^2 > b^2 + b^2$
$c^2 > 2b^2$.

(D) The maximum value of the quadratic expression $1-2x-5x^2$ is found at $x = -\frac{b}{2a} = -\frac{-2}{2(-5)} = -\frac{1}{5}$.
Substituting $x = -\frac{1}{5}$,we get $a = 1 - 2(-\frac{1}{5}) - 5(-\frac{1}{5})^2 = 1 + \frac{2}{5} - \frac{1}{5} = \frac{6}{5}$.
The minimum value of the quadratic expression $x^2-2x+5$ is found at $x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1$.
Substituting $x = 1$,we get $b = (1)^2 - 2(1) + 5 = 4$.
Now,substitute $a = \frac{6}{5}$ and $b = 4$ into the expression $5ax^2+bx+7$:
$5(\frac{6}{5})x^2 + 4x + 7 = 6x^2 + 4x + 7$.
For the quadratic $6x^2 + 4x + 7 > 0$,we check the discriminant $D = b^2 - 4ac = (4)^2 - 4(6)(7) = 16 - 168 = -152$.
Since $D < 0$ and the leading coefficient $6 > 0$,the expression $6x^2 + 4x + 7$ is always positive for all real values of $x$.
Thus,the set of all values of $x$ is $(-\infty, \infty)$.

(A) For a quadratic expression $f(x) = ax^2+bx+c$,the minimum value occurs at $x = -\frac{b}{2a}$.
Here,$f(x) = x^2+5x-2$,so $a=1, b=5, c=-2$.
The minimum value occurs at $a = -\frac{5}{2(1)} = -2.5$.
The minimum value $M$ is $f(-2.5) = (-2.5)^2 + 5(-2.5) - 2 = 6.25 - 12.5 - 2 = -8.25$.
Now,$\frac{M}{a} = \frac{-8.25}{-2.5} = 3.3$.

(D) For $f(x) = x^2 - 2(4K - 1)x + g(K) > 0$ to hold for all $x \in R$,the discriminant $D$ must be less than $0$.
$D = [-2(4K - 1)]^2 - 4(1)(g(K)) < 0$
$4(16K^2 - 8K + 1) - 4(15K^2 - 2K - 7) < 0$
$16K^2 - 8K + 1 - 15K^2 + 2K + 7 < 0$
$K^2 - 6K + 8 < 0$
$(K - 2)(K - 4) < 0$
Thus,$K \in (2, 4)$,so $a = 2$ and $b = 4$.
The function $g(K) = 15K^2 - 2K - 7$ is an upward-opening parabola with its vertex at $K = -(-2) / (2 \times 15) = 1/15$.
Since $1/15 \notin (2, 4)$,the function $g(K)$ is strictly increasing on the interval $(2, 4)$.
Therefore,$g(K)$ does not attain a maximum or a minimum value within the open interval $(2, 4)$.

(C) Given $f(x)=ax^2-bx-a$ is a quadratic expression such that $f(x) \leq K, \forall x \in R$.
This implies $ax^2-bx-a-K \leq 0, \forall x \in R$.
For a quadratic expression $Ax^2+Bx+C$ to be less than or equal to $0$ for all $x$,the coefficient of $x^2$ must be negative $(a < 0)$ and the discriminant $D$ must be less than or equal to $0$.
Here,$D = (-b)^2 - 4(a)(-a-K) \leq 0$.
$b^2 + 4a(a+K) \leq 0$.
$b^2 + 4a^2 + 4aK \leq 0$.
Since $b^2 + 4a^2 \geq 0$,for the expression to be $\leq 0$,$4aK$ must be negative.
Given $a < 0$,for $4aK < 0$ to hold,$K$ must be positive.
Thus,$K > 0$.

(D) Let $f(x) = -ax^2 + 2x - 7$. For the function to have a maximum value,we must have $a > 0$.
Completing the square:
$f(x) = -a(x^2 - \frac{2}{a}x) - 7$
$f(x) = -a(x^2 - \frac{2}{a}x + \frac{1}{a^2} - \frac{1}{a^2}) - 7$
$f(x) = -a(x - \frac{1}{a})^2 + \frac{1}{a} - 7$
The maximum value of $f(x)$ is $\frac{1}{a} - 7$.
Given that the maximum value cannot exceed $20$,we have:
$\frac{1}{a} - 7 \leq 20$
$\frac{1}{a} \leq 27$
Since $a > 0$,we get $a \geq \frac{1}{27}$.
Thus,the minimum value of $a$ is $\frac{1}{27}$.

(B) Let $f(x) = x^2 - 4ax + (4a^2 - 3a + 1)$. For both roots to be greater than $1$,the following conditions must be satisfied:
$1$. Discriminant $D \geq 0$:
$D = (-4a)^2 - 4(1)(4a^2 - 3a + 1) = 16a^2 - 16a^2 + 12a - 4 = 12a - 4 \geq 0 \Rightarrow a \geq \frac{1}{3}$.
$2$. Vertex position: $\frac{-b}{2a} > 1$:
$\frac{4a}{2} > 1$ $\Rightarrow 2a > 1$ $\Rightarrow a > \frac{1}{2}$.
$3$. $f(1) > 0$:
$f(1) = 1^2 - 4a(1) + 4a^2 - 3a + 1 = 4a^2 - 7a + 2 > 0$.
Roots of $4a^2 - 7a + 2 = 0$ are $a = \frac{7 \pm \sqrt{49 - 32}}{8} = \frac{7 \pm \sqrt{17}}{8}$.
Since the parabola opens upward,$4a^2 - 7a + 2 > 0$ for $a \in \left(-\infty, \frac{7-\sqrt{17}}{8}\right) \cup \left(\frac{7+\sqrt{17}}{8}, \infty\right)$.
Taking the intersection of $a \geq \frac{1}{3}$,$a > \frac{1}{2}$,and $a \in \left(-\infty, \frac{7-\sqrt{17}}{8}\right) \cup \left(\frac{7+\sqrt{17}}{8}, \infty\right)$:
Since $\frac{7+\sqrt{17}}{8} \approx \frac{7+4.12}{8} \approx 1.39 > \frac{1}{2}$,the intersection is $a \in \left(\frac{7+\sqrt{17}}{8}, \infty\right)$.

(B) Let $f(x) = x^2+x+a$. For the roots to exceed $a$,the following conditions must be satisfied:
$1$. $D \geq 0$ $\Rightarrow 1-4a \geq 0$ $\Rightarrow a \leq \frac{1}{4}$.
$2$. $f(a) > 0$ $\Rightarrow a^2+a+a > 0$ $\Rightarrow a^2+2a > 0$ $\Rightarrow a(a+2) > 0$ $\Rightarrow a \in (-\infty, -2) \cup (0, \infty)$.
$3$. $-\frac{b}{2a} > a$ $\Rightarrow -\frac{1}{2} > a$ $\Rightarrow a < -\frac{1}{2}$.
Taking the intersection of all three conditions:
$a \in (-\infty, \frac{1}{4}] \cap ((-\infty, -2) \cup (0, \infty)) \cap (-\infty, -\frac{1}{2})$
$= (-\infty, -2)$.

(D) Let $f(x) = a x^{2} + b x + c$.
Given that $f(1) = a + b + c < 0$.
Since the quadratic equation $a x^{2} + b x + c = 0$ has imaginary roots,the graph of $f(x)$ does not intersect the $x$-axis.
This means $f(x)$ is either always positive $(a > 0)$ or always negative $(a < 0)$.
Since $f(1) < 0$,$f(x)$ must be always negative,which implies $a < 0$.
Also,$f(0) = c$. Since $f(x)$ is always negative,$f(0) < 0$,which implies $c < 0$.
Thus,$a < 0$ and $c < 0$.

(C) Given the quadratic expression is $(2x+1)^{2} - px + q \neq 0$ for all real $x$.
Expanding the expression:
$4x^{2} + 4x + 1 - px + q \neq 0$
$4x^{2} + (4-p)x + (1+q) \neq 0$
For a quadratic expression $ax^{2} + bx + c$ to be non-zero for all real $x$,its discriminant $D$ must be less than $0$.
$D = b^{2} - 4ac < 0$
Substituting the coefficients $a = 4$,$b = (4-p)$,and $c = (1+q)$:
$(4-p)^{2} - 4(4)(1+q) < 0$
$16 - 8p + p^{2} - 16 - 16q < 0$
$p^{2} - 8p - 16q < 0$

(C) Let $f(x) = ax^2+bx+c$. Since $a > 0$,the parabola opens upwards.
Given that the roots $\alpha$ and $\beta$ satisfy $\alpha < -2$ and $\beta > 2$,the value of the function at $x = -1$ must be negative because $-1$ lies between the roots $\alpha$ and $\beta$ (since $\alpha < -2 < -1 < 2 < \beta$).
Thus,$f(-1) < 0$.
Substituting $x = -1$ into the quadratic expression,we get $f(-1) = a(-1)^2 + b(-1) + c = a - b + c$.
Therefore,$a - b + c < 0$.

(D) Let $f(x) = x^2+px-q^2$.
Since the coefficient of $x^2$ is $1 > 0$,the parabola opens upwards.
Given that one root $\alpha < 2$ and the other root $\beta > 2$,the value of the function at $x=2$ must be negative,i.e.,$f(2) < 0$.
Substituting $x=2$ into the quadratic expression:
$f(2) = (2)^2 + p(2) - q^2 < 0$
$4 + 2p - q^2 < 0$
Thus,the correct option is $D$.

(D) Let $f(x) = x^{2}-3x+k$.
For the roots to lie in the interval $(0,1)$,the vertex of the parabola must lie within $(0,1)$.
The $x$-coordinate of the vertex is given by $x = -\frac{b}{2a} = -\frac{-3}{2(1)} = \frac{3}{2} = 1.5$.
Since $1.5 \notin (0,1)$,it is impossible for both roots to lie in the interval $(0,1)$ simultaneously.
Therefore,there is no value of $k$ that satisfies the given condition.

(D) Given the quadratic equation $f(x) = x^{2}+2ax+(3a+10) = 0$.
Since the roots $\alpha$ and $\beta$ satisfy $\alpha < 1 < \beta$,the value of the function at $x=1$ must be negative,i.e.,$f(1) < 0$.
Substituting $x=1$ into the equation:
$f(1) = (1)^{2} + 2a(1) + (3a+10) < 0$
$1 + 2a + 3a + 10 < 0$
$5a + 11 < 0$
$5a < -11$
$a < -11/5$
Therefore,the set of all possible values of $a$ is $(-\infty, -11/5)$.