Mix Examples - Quadratic Equations Questions in English

(A) To determine if an equation is quadratic,we simplify it to the standard form $ax^{2}+bx+c=0$,where $a \neq 0$.
Check option $(A)$: $(x+2)(x-1) = x^{2}-x+2x-2 = x^{2}+x-2$. Given equation: $x^{2}+x-2 = x^{2}-2x-3$. Simplifying,$3x+1=0$. This is a linear equation,not a quadratic equation.
Check option $(B)$: $(x+2)^{2} = 2(x+3) \implies x^{2}+4x+4 = 2x+6 \implies x^{2}+2x-2=0$. This is a quadratic equation.
Check option $(C)$: $x^{2}+3x = (-1)(1-3x)^{2} \implies x^{2}+3x = -(1-6x+9x^{2}) \implies x^{2}+3x = -1+6x-9x^{2} \implies 10x^{2}-3x+1=0$. This is a quadratic equation.
Check option $(D)$: $x^{3}-x^{2}+2x+1 = (x+1)^{3} \implies x^{3}-x^{2}+2x+1 = x^{3}+3x^{2}+3x+1 \implies -x^{2}+2x = 3x^{2}+3x \implies 4x^{2}+x=0$. This is a quadratic equation.
Therefore,the equation in option $(A)$ is not a quadratic equation.

(B) To solve the quadratic equation $4x^{2} - \sqrt{3}x - 5 = 0$ by the method of completing the square,we first ensure the coefficient of $x^{2}$ is $1$ or a perfect square. Here,$4x^{2} = (2x)^{2}$.
We rewrite the equation as $(2x)^{2} - 2(2x)(\frac{\sqrt{3}}{4}) - 5 = 0$.
To complete the square in the form $(a - b)^{2} = a^{2} - 2ab + b^{2}$,we identify $a = 2x$ and $b = \frac{\sqrt{3}}{4}$.
The term to be added and subtracted is $b^{2} = (\frac{\sqrt{3}}{4})^{2} = \frac{3}{16}$.
Thus,the constant is $\frac{3}{16}$.

(C) For an equation to be a quadratic equation,it must be in the form $ax^{2}+bx+c=0$,where $a \neq 0$.
$(A)$ $x^{2}+2x+1 = (4-x)^{2}+3$
$\Rightarrow x^{2}+2x+1 = 16-8x+x^{2}+3$
$\Rightarrow 10x-18=0$. This is a linear equation.
$(B)$ $-2x^{2} = (5-x)(2x-\frac{2}{5})$
$\Rightarrow -2x^{2} = 10x-2-2x^{2}+\frac{2x}{5}$
$\Rightarrow 0 = \frac{52x}{5}-2$. This is a linear equation.
$(C)$ $x^{3}-x^{2} = (x-1)^{3}$
$\Rightarrow x^{3}-x^{2} = x^{3}-3x^{2}+3x-1$
$\Rightarrow 2x^{2}-3x+1=0$. This is a quadratic equation as it is in the form $ax^{2}+bx+c=0$ with $a=2 \neq 0$.
$(D)$ $(k+1)x^{2}+\frac{3}{2}x=7$,where $k=-1$
$\Rightarrow (-1+1)x^{2}+\frac{3}{2}x=7$
$\Rightarrow \frac{3}{2}x-7=0$. This is a linear equation.

(D) quadratic equation is of the form $ax^{2}+bx+c=0$,where $a \neq 0$.
$(A)$ $2(x-1)^{2}=4 x^{2}-2 x+1 \Rightarrow 2(x^{2}-2x+1)=4x^{2}-2x+1 \Rightarrow 2x^{2}-4x+2=4x^{2}-2x+1 \Rightarrow 2x^{2}+2x-1=0$. This is a quadratic equation.
$(B)$ $2 x-x^{2}=x^{2}+5 \Rightarrow 2x^{2}-2x+5=0$. This is a quadratic equation.
$(C)$ $(-\sqrt{2} x+\sqrt{3})^{2}=3 x^{2}-5 x \Rightarrow 2x^{2}-2\sqrt{6}x+3=3x^{2}-5x \Rightarrow x^{2}-(5-2\sqrt{6})x-3=0$. This is a quadratic equation.
$(D)$ $(x^{2}+2 x)^{2}=x^{4}+3+4 x^{2} \Rightarrow x^{4}+4x^{3}+4x^{2}=x^{4}+3+4x^{2} \Rightarrow 4x^{3}-3=0$. Since the highest power of $x$ is $3$,this is a cubic equation,not a quadratic equation.

(A) To determine if $x = 2$ is a root of an equation,we substitute $x = 2$ into each equation and check if the result is $0$.
$(A)$ For $2x^2 - 7x + 6 = 0$:
$2(2)^2 - 7(2) + 6 = 2(4) - 14 + 6 = 8 - 14 + 6 = 0$.
Since the result is $0$,$x = 2$ is a root of this equation.
$(B)$ For $x^2 + 3x - 12 = 0$:
$(2)^2 + 3(2) - 12 = 4 + 6 - 12 = -2 \neq 0$.
So,$x = 2$ is not a root.
$(C)$ For $x^2 - 4x + 5 = 0$:
$(2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1 \neq 0$.
So,$x = 2$ is not a root.
$(D)$ For $3x^2 - 6x - 2 = 0$:
$3(2)^2 - 6(2) - 2 = 12 - 12 - 2 = -2 \neq 0$.
So,$x = 2$ is not a root.
Therefore,the correct equation is $2x^2 - 7x + 6 = 0$.

(B) Since $\frac{1}{2}$ is a root of the quadratic equation $x^{2}+kx-\frac{5}{4}=0$,it must satisfy the equation.
Substituting $x = \frac{1}{2}$ into the equation:
$(\frac{1}{2})^{2} + k(\frac{1}{2}) - \frac{5}{4} = 0$
$\frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0$
$\frac{1 + 2k - 5}{4} = 0$
$2k - 4 = 0$
$2k = 4$
$k = 2$

(C) For a quadratic equation $ax^2 + bx + c = 0$,the sum of the roots is given by $\frac{-b}{a}$.
$(A)$ For $2x^2 - 3x + 6 = 0$,$a=2, b=-3$. Sum of roots $= -(-3)/2 = 3/2 = 1.5$.
$(B)$ For $\sqrt{2}x^2 - \frac{3}{\sqrt{2}}x + 1 = 0$,multiplying by $\sqrt{2}$ gives $2x^2 - 3x + \sqrt{2} = 0$. Here $a=2, b=-3$. Sum of roots $= -(-3)/2 = 3/2 = 1.5$.
$(C)$ For $-x^2 + 3x - 3 = 0$,$a=-1, b=3$. Sum of roots $= -(3)/(-1) = 3$.
$(D)$ For $3x^2 - 3x + 3 = 0$,dividing by $3$ gives $x^2 - x + 1 = 0$. Here $a=1, b=-1$. Sum of roots $= -(-1)/1 = 1$.
Thus,the equation with the sum of roots equal to $3$ is $-x^2 + 3x - 3 = 0$.

(D) The given quadratic equation is $2x^{2}-kx+k=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=2$,$b=-k$,and $c=k$.
For a quadratic equation to have equal roots,the discriminant $D$ must be equal to $0$.
$D = b^{2}-4ac = 0$
Substituting the values,we get $(-k)^{2}-4(2)(k) = 0$.
$k^{2}-8k = 0$
Factoring out $k$,we get $k(k-8) = 0$.
Therefore,$k=0$ or $k=8$.
Thus,the required values of $k$ are $0$ and $8$.

(A) The given quadratic equation is $9x^{2} + \frac{3}{4}x - \sqrt{2} = 0$.
To complete the square,we first express the equation in the form $(ax)^{2} + 2(ax)(b) + b^{2}$.
We can rewrite the equation as $(3x)^{2} + 2(3x)(\frac{1}{8}) - \sqrt{2} = 0$.
Here,the term $b$ is $\frac{1}{8}$.
To complete the square,we need to add and subtract $b^{2} = (\frac{1}{8})^{2} = \frac{1}{64}$.
Thus,the constant that must be added and subtracted is $\frac{1}{64}$.

(B) Given equation is $2x^{2} - \sqrt{5}x + 1 = 0$.
On comparing with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 2, b = -\sqrt{5}, c = 1$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values:
$D = (-\sqrt{5})^{2} - 4(2)(1) = 5 - 8 = -3$.
Since the discriminant $D < 0$,the quadratic equation has no real roots (i.e.,it has imaginary roots).

(C) For a quadratic equation $ax^{2}+bx+c=0$,the roots are distinct and real if the discriminant $D = b^{2}-4ac > 0$.
$(a)$ For $2x^{2}-3\sqrt{2}x+\frac{9}{4}=0$,$a=2, b=-3\sqrt{2}, c=\frac{9}{4}$.
$D = (-3\sqrt{2})^{2}-4(2)(\frac{9}{4}) = 18-18 = 0$. Since $D=0$,the roots are real and equal.
$(b)$ For $x^{2}+3x+2\sqrt{2}=0$,$a=1, b=3, c=2\sqrt{2}$.
$D = (3)^{2}-4(1)(2\sqrt{2}) = 9-8\sqrt{2} < 0$. Since $D < 0$,the roots are not real.
$(c)$ For $x^{2}+x-5=0$,$a=1, b=1, c=-5$.
$D = (1)^{2}-4(1)(-5) = 1+20 = 21$. Since $D>0$,the roots are distinct and real.
$(d)$ For $5x^{2}-3x+1=0$,$a=5, b=-3, c=1$.
$D = (-3)^{2}-4(5)(1) = 9-20 = -11 < 0$. Since $D < 0$,the roots are not real.
Thus,the correct option is $(c)$.

(D) To determine if a quadratic equation $ax^2 + bx + c = 0$ has real roots,we calculate the discriminant $D = b^2 - 4ac$. If $D < 0$,the equation has no real roots.
$A)$ For $3x^2 + 4\sqrt{3}x + 4 = 0$,$D = (4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$. Since $D = 0$,it has real and equal roots.
$B)$ For $x^2 - 4x - 3\sqrt{2} = 0$,$D = (-4)^2 - 4(1)(-3\sqrt{2}) = 16 + 12\sqrt{2} > 0$. It has real and distinct roots.
$C)$ For $x^2 + 4x - 3\sqrt{2} = 0$,$D = (4)^2 - 4(1)(-3\sqrt{2}) = 16 + 12\sqrt{2} > 0$. It has real and distinct roots.
$D)$ For $x^2 - 4x + 3\sqrt{2} = 0$,$D = (-4)^2 - 4(1)(3\sqrt{2}) = 16 - 12\sqrt{2}$. Since $\sqrt{2} \approx 1.414$,$12\sqrt{2} \approx 16.968$. Thus,$D = 16 - 16.968 = -0.968 < 0$. Therefore,this equation has no real roots.

(A) Given equation is $(x^{2}+1)^{2}-x^{2}=0$.
Expanding the square: $x^{4}+1+2x^{2}-x^{2}=0$,which simplifies to $x^{4}+x^{2}+1=0$.
Let $x^{2}=y$. Since $x$ is a real number,$x^{2} \ge 0$,so $y \ge 0$.
The equation becomes $y^{2}+y+1=0$.
Comparing with $ay^{2}+by+c=0$,we have $a=1, b=1, c=1$.
The discriminant $D = b^{2}-4ac = (1)^{2}-4(1)(1) = 1-4 = -3$.
Since $D < 0$,the quadratic equation in $y$ has no real roots.
Therefore,the original equation $(x^{2}+1)^{2}-x^{2}=0$ has no real roots.

(NO) To determine if the equation has real roots,we first simplify it:
$(x-1)^{2}+2(x+1)=0$
$(x^{2}-2x+1)+2x+2=0$
$x^{2}+3=0$
Now,we compare this with the standard quadratic form $ax^{2}+bx+c=0$,where $a=1, b=0, c=3$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (0)^{2}-4(1)(3) = -12$.
Since the discriminant $D < 0$,the equation has no real roots.

(B) The statement is False.
Consider the general form of a quadratic equation: $ax^2 + bx + c = 0$.
If the coefficient of $x$ is zero,then $b = 0$,and the equation becomes $ax^2 + c = 0$.
The roots of this equation are given by $x^2 = -c/a$,which implies $x = \pm \sqrt{-c/a}$.
For the roots to be real,we require $-c/a \geq 0$,which means $c/a \leq 0$.
This occurs if $a$ and $c$ have opposite signs or if $c = 0$.
For example,if $x^2 - 4 = 0$,then $x^2 = 4$,so $x = \pm 2$,which are real roots.
Thus,the statement is false.

(N/A) Given equation is $x^{2}-3x+4=0$.
On comparing this with the standard form $ax^{2}+bx+c=0$,we get:
$a=1, b=-3, c=4$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
Substituting the values:
$D = (-3)^{2} - 4(1)(4)$
$D = 9 - 16$
$D = -7$.
Since $D < 0$,the quadratic equation $x^{2}-3x+4=0$ has no real roots. Therefore,it does not have two distinct real roots.

(A) The given quadratic equation is $2 x^{2}+x-1=0$.
Comparing this with the standard form $a x^{2}+b x+c=0$,we get $a=2$,$b=1$,and $c=-1$.
The discriminant $D$ is given by the formula $D = b^{2}-4 a c$.
Substituting the values,we get $D = (1)^{2}-4(2)(-1) = 1+8 = 9$.
Since $D > 0$,the quadratic equation has two distinct real roots.

(B) The given quadratic equation is $2 x^{2}-6 x+\frac{9}{2}=0$.
Comparing this with the standard form $a x^{2}+b x+c=0$,we get:
$a=2, b=-6, c=\frac{9}{2}$.
To determine the nature of the roots,we calculate the discriminant $D = b^{2}-4ac$:
$D = (-6)^{2} - 4(2)(\frac{9}{2})$
$D = 36 - 36 = 0$.
Since the discriminant $D = 0$,the quadratic equation has two equal real roots. Therefore,it does not have two distinct real roots.

(A) The given quadratic equation is $3 x^{2}-4 x+1=0$.
Comparing this with the standard form $a x^{2}+b x+c=0$,we get:
$a=3, b=-4, c=1$.
The discriminant $D$ is given by the formula $D = b^{2}-4ac$.
Substituting the values,we get:
$D = (-4)^{2} - 4(3)(1)$
$D = 16 - 12$
$D = 4$.
Since $D > 0$,the quadratic equation has two distinct real roots.

(NO) Given equation is $(x+4)^{2}-8x=0$.
Expanding the term: $x^{2}+16+8x-8x=0$ (using $(a+b)^{2}=a^{2}+2ab+b^{2}$).
Simplifying: $x^{2}+16=0$.
Writing in standard form $ax^{2}+bx+c=0$: $x^{2}+0x+16=0$.
Comparing with $ax^{2}+bx+c=0$,we get $a=1, b=0, c=16$.
The discriminant $D$ is given by $D=b^{2}-4ac$.
Substituting the values: $D=(0)^{2}-4(1)(16) = 0-64 = -64$.
Since $D < 0$,the quadratic equation has no real roots (it has two imaginary roots).

(A) Given equation is $(x-\sqrt{2})^{2}-2(x+1)=0$.
Expanding the equation: $x^{2}-2\sqrt{2}x+2-2x-2=0$.
Simplifying: $x^{2}-(2+2\sqrt{2})x=0$.
Comparing with $ax^{2}+bx+c=0$,we get $a=1, b=-(2+2\sqrt{2}), c=0$.
The discriminant $D = b^{2}-4ac = (-(2+2\sqrt{2}))^{2}-4(1)(0)$.
$D = (2+2\sqrt{2})^{2} = 4 + 8 + 8\sqrt{2} = 12+8\sqrt{2}$.
Since $D > 0$,the quadratic equation has two distinct real roots.

(A) Given,the quadratic equation is $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0$.
On comparing this with the standard form $ax^{2}+bx+c=0$,we get:
$a=\sqrt{2}$,$b=-\frac{3}{\sqrt{2}}$,and $c=\frac{1}{\sqrt{2}}$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
Substituting the values:
$D = \left(-\frac{3}{\sqrt{2}}\right)^{2} - 4(\sqrt{2})\left(\frac{1}{\sqrt{2}}\right)$
$D = \frac{9}{2} - 4$
$D = \frac{9-8}{2} = \frac{1}{2}$.
Since $D = \frac{1}{2} > 0$,the discriminant is positive.
Therefore,the quadratic equation $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0$ has two distinct real roots.

(NO) Given equation is $x(1-x)-2=0$.
Expanding the terms,we get $x-x^{2}-2=0$.
Rearranging the equation in the standard form $ax^{2}+bx+c=0$,we get $x^{2}-x+2=0$.
Comparing this with $ax^{2}+bx+c=0$,we have $a=1, b=-1, c=2$.
The discriminant $D$ is given by $D=b^{2}-4ac$.
Substituting the values,$D=(-1)^{2}-4(1)(2) = 1-8 = -7$.
Since $D < 0$,the quadratic equation $x^{2}-x+2=0$ has no real roots (it has two distinct imaginary roots).

(A) Given equation is $(x-1)(x+2)+2=0$.
Expanding the terms,we get $x^{2} + 2x - x - 2 + 2 = 0$.
Simplifying,we get $x^{2} + x = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we have $a = 1, b = 1, c = 0$.
The discriminant $D$ is given by $D = b^{2} - 4ac$.
Substituting the values,$D = (1)^{2} - 4(1)(0) = 1 - 0 = 1$.
Since $D > 0$,the quadratic equation has two distinct real roots.

(A) Given equation is $(x+1)(x-2)+x=0$.
Expanding the terms:
$x^2 - 2x + x - 2 + x = 0$
Simplifying the equation:
$x^2 - 2 = 0$
Comparing this with the standard form $ax^2 + bx + c = 0$,we get:
$a = 1, b = 0, c = -2$
The discriminant $D$ is given by $D = b^2 - 4ac$.
Substituting the values:
$D = (0)^2 - 4(1)(-2) = 0 + 8 = 8$.
Since $D > 0$,the quadratic equation has two distinct real roots.

(N/A) $(i)$ False. $A$ quadratic equation of the form $ax^{2} + bx + c = 0$ (where $a \neq 0$) has exactly two roots,which may be real or complex.
$(ii)$ False. $A$ quadratic equation does not necessarily have a real root. For example,the equation $x^{2} + 4 = 0$ has no real roots because $x^{2} = -4$,and the square of a real number cannot be negative.

(A) $(i)$ True. $A$ quadratic equation is of the form $ax^{2} + bx + c = 0$ where $a \neq 0$. By the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ roots. Since a quadratic equation has degree $2$,it can have at most $2$ roots.
$(ii)$ True. For a quadratic equation $ax^{2} + bx + c = 0$,the discriminant is $D = b^{2} - 4ac$. If the coefficient of $x^{2}$ $(a)$ and the constant term $(c)$ have opposite signs,then $ac < 0$. Consequently,$-4ac > 0$. Since $b^{2} \geq 0$,it follows that $D = b^{2} - 4ac > 0$. Because the discriminant is positive,the quadratic equation must have two distinct real roots.

(A) $(i)$ True. For a quadratic equation $ax^{2} + bx + c = 0$,the discriminant is $D = b^{2} - 4ac$. Given $b = 0$,we have $D = -4ac$. If $a$ and $c$ have the same sign,then $ac > 0$,which implies $D = -4ac < 0$. Since the discriminant is negative,the equation has no real roots.
$(ii)$ False. $A$ quadratic equation has exactly two roots (which may be real and distinct,real and equal,or complex/imaginary). It does not necessarily have "at least" two roots in the context of real numbers,and it cannot have more than two roots.

(B) The statement is false. $A$ quadratic equation with integral coefficients does not necessarily have integral roots.
Consider the quadratic equation $2x^2 + x - 6 = 0$, where the coefficients $2, 1, \text{and } -6$ are integers.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:
$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)}$
$x = \frac{-1 \pm \sqrt{1 + 48}}{4}$
$x = \frac{-1 \pm \sqrt{49}}{4}$
$x = \frac{-1 \pm 7}{4}$
This gives two roots: $x_1 = \frac{6}{4} = 1.5$ and $x_2 = \frac{-8}{4} = -2$.
Since $1.5$ is not an integer, the claim that a quadratic equation with integral coefficients must have integral roots is incorrect.

(A) Yes,such a quadratic equation exists. Consider the quadratic equation $2x^2 + x - 4 = 0$. Here,the coefficients $2, 1,$ and $-4$ are rational numbers.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-4)}}{2(2)}$
$x = \frac{-1 \pm \sqrt{1 + 32}}{4}$
$x = \frac{-1 \pm \sqrt{33}}{4}$
The roots are $\frac{-1 + \sqrt{33}}{4}$ and $\frac{-1 - \sqrt{33}}{4}$. Since $\sqrt{33}$ is an irrational number,both roots are irrational.

(C) Yes,such a quadratic equation exists.
Consider the quadratic equation $\sqrt{3}x^2 - 7\sqrt{3}x + 12\sqrt{3} = 0$.
Here,the coefficients are $a = \sqrt{3}$,$b = -7\sqrt{3}$,and $c = 12\sqrt{3}$,which are all distinct irrational numbers.
Dividing the entire equation by $\sqrt{3}$,we get $x^2 - 7x + 12 = 0$.
Factoring the quadratic,we get $(x - 3)(x - 4) = 0$.
Thus,the roots are $x = 3$ and $x = 4$,which are both rational numbers.

(B) To determine if $0.2$ is a root of the equation $x^{2}-0.4=0$,we substitute $x = 0.2$ into the equation.
Substituting $x = 0.2$ into the left-hand side $(LHS)$ gives:
$(0.2)^{2} - 0.4 = 0.04 - 0.4 = -0.36$.
Since the result is $-0.36$ and not $0$,the value $0.2$ does not satisfy the equation.
Therefore,$0.2$ is not a root of the equation $x^{2}-0.4=0$.

(A) Given the quadratic equation: $x^{2}+bx+c=0$ ....$(i)$
Substitute $b=0$ into equation $(i)$:
$x^{2}+0(x)+c=0$
$x^{2}+c=0$
$x^{2}=-c$
Since it is given that $c < 0$,let $c = -k$ where $k > 0$. Then:
$x^{2} = -(-k) = k$
$x = \pm \sqrt{k}$
Thus,the roots are $\sqrt{k}$ and $-\sqrt{k}$.
These roots are numerically equal (both have magnitude $\sqrt{k}$) and opposite in sign (one is positive and one is negative). Therefore,the statement is true.

(B) For the quadratic equation $ax^{2}+bx+c=0$,the roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Here,$a=2$,$b=-\sqrt{5}$,and $c=-2$.
First,calculate the discriminant $D = b^{2}-4ac$:
$D = (-\sqrt{5})^{2} - 4 \times 2 \times (-2) = 5 + 16 = 21$.
Now,substitute the values into the quadratic formula:
$x = \frac{-(-\sqrt{5}) \pm \sqrt{21}}{2 \times 2} = \frac{\sqrt{5} \pm \sqrt{21}}{4}$.
Thus,the roots are $\frac{\sqrt{5}+\sqrt{21}}{4}$ and $\frac{\sqrt{5}-\sqrt{21}}{4}$.

(D) To find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$,we factorize the quadratic polynomial.
We need two numbers whose product is $6 \times (-2) = -12$ and whose sum is $-\sqrt{2}$.
These numbers are $-3\sqrt{2}$ and $2\sqrt{2}$.
$6x^{2} - 3\sqrt{2}x + 2\sqrt{2}x - 2 = 0$
$3x(2x - \sqrt{2}) + \sqrt{2}(2x - \sqrt{2}) = 0$
$(3x + \sqrt{2})(2x - \sqrt{2}) = 0$
Setting each factor to zero:
$3x + \sqrt{2} = 0 \implies x = -\frac{\sqrt{2}}{3}$
$2x - \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{2}$
Thus,the roots are $-\frac{\sqrt{2}}{3}$ and $\frac{\sqrt{2}}{2}$.

(D) The given equation is $2x^{2} - 3x - 5 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 2, b = -3, c = -5$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Substituting the values:
$x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(2)(-5)}}{2(2)}$
$x = \frac{3 \pm \sqrt{9 + 40}}{4}$
$x = \frac{3 \pm \sqrt{49}}{4}$
$x = \frac{3 \pm 7}{4}$.
Calculating the two roots:
$x_{1} = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}$
$x_{2} = \frac{3 - 7}{4} = \frac{-4}{4} = -1$.
Thus,the roots of the equation are $\frac{5}{2}$ and $-1$.

(A) Given equation is $5x^{2} + 13x + 8 = 0$.
On comparing with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 5, b = 13, c = 8$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Substituting the values:
$x = \frac{-(13) \pm \sqrt{(13)^{2} - 4(5)(8)}}{2(5)}$
$x = \frac{-13 \pm \sqrt{169 - 160}}{10}$
$x = \frac{-13 \pm \sqrt{9}}{10}$
$x = \frac{-13 \pm 3}{10}$.
Case $1$: $x = \frac{-13 + 3}{10} = \frac{-10}{10} = -1$.
Case $2$: $x = \frac{-13 - 3}{10} = \frac{-16}{10} = -\frac{8}{5}$.
Thus,the roots are $-1$ and $-\frac{8}{5}$.

(B) The given equation is $-3x^{2} + 5x + 12 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = -3$,$b = 5$,and $c = 12$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Substituting the values,we get $x = \frac{-(5) \pm \sqrt{(5)^{2} - 4(-3)(12)}}{2(-3)}$.
Simplifying the expression,$x = \frac{-5 \pm \sqrt{25 + 144}}{-6} = \frac{-5 \pm \sqrt{169}}{-6}$.
Since $\sqrt{169} = 13$,we have $x = \frac{-5 \pm 13}{-6}$.
Taking the positive sign: $x = \frac{-5 + 13}{-6} = \frac{8}{-6} = -\frac{4}{3}$.
Taking the negative sign: $x = \frac{-5 - 13}{-6} = \frac{-18}{-6} = 3$.
Thus,the roots of the equation are $-\frac{4}{3}$ and $3$.

(C) Given equation is $-x^{2}+7x-10=0$.
On comparing with the standard form $ax^{2}+bx+c=0$,we get:
$a = -1, b = 7, c = -10$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values:
$x = \frac{-(7) \pm \sqrt{(7)^{2}-4(-1)(-10)}}{2(-1)}$
$x = \frac{-7 \pm \sqrt{49-40}}{-2}$
$x = \frac{-7 \pm \sqrt{9}}{-2}$
$x = \frac{-7 \pm 3}{-2}$.
Calculating the two roots:
$x_1 = \frac{-7 + 3}{-2} = \frac{-4}{-2} = 2$
$x_2 = \frac{-7 - 3}{-2} = \frac{-10}{-2} = 5$.
Thus,the roots of the equation are $2$ and $5$.

(D) The given equation is $x^{2}+2 \sqrt{2} x-6=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=2 \sqrt{2}$,and $c=-6$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values,we get $x = \frac{-(2 \sqrt{2}) \pm \sqrt{(2 \sqrt{2})^{2}-4(1)(-6)}}{2(1)}$.
$x = \frac{-2 \sqrt{2} \pm \sqrt{8+24}}{2} = \frac{-2 \sqrt{2} \pm \sqrt{32}}{2}$.
Since $\sqrt{32} = 4 \sqrt{2}$,we have $x = \frac{-2 \sqrt{2} \pm 4 \sqrt{2}}{2}$.
Calculating the two roots:
$x_1 = \frac{-2 \sqrt{2} + 4 \sqrt{2}}{2} = \frac{2 \sqrt{2}}{2} = \sqrt{2}$.
$x_2 = \frac{-2 \sqrt{2} - 4 \sqrt{2}}{2} = \frac{-6 \sqrt{2}}{2} = -3 \sqrt{2}$.
Thus,the roots are $\sqrt{2}$ and $-3 \sqrt{2}$.

(A) The given quadratic equation is $x^{2}-3 \sqrt{5} x+10=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get:
$a=1, b=-3 \sqrt{5}, c=10$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values:
$x = \frac{-(-3 \sqrt{5}) \pm \sqrt{(-3 \sqrt{5})^{2}-4(1)(10)}}{2(1)}$
$x = \frac{3 \sqrt{5} \pm \sqrt{45-40}}{2}$
$x = \frac{3 \sqrt{5} \pm \sqrt{5}}{2}$
Case $1$: $x = \frac{3 \sqrt{5} + \sqrt{5}}{2} = \frac{4 \sqrt{5}}{2} = 2 \sqrt{5}$.
Case $2$: $x = \frac{3 \sqrt{5} - \sqrt{5}}{2} = \frac{2 \sqrt{5}}{2} = \sqrt{5}$.
Thus,the roots of the equation are $2 \sqrt{5}$ and $\sqrt{5}$.

(B) Given equation is $\frac{1}{2} x^{2}-\sqrt{11} x+1=0$.
On comparing with $ax^{2}+bx+c=0$,we get:
$a=\frac{1}{2}, b=-\sqrt{11}, c=1$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values:
$x = \frac{-(-\sqrt{11}) \pm \sqrt{(-\sqrt{11})^{2}-4 \times \frac{1}{2} \times 1}}{2 \times \frac{1}{2}}$
$x = \frac{\sqrt{11} \pm \sqrt{11-2}}{1}$
$x = \sqrt{11} \pm \sqrt{9}$
$x = \sqrt{11} \pm 3$.
Therefore,the roots are $3+\sqrt{11}$ and $\sqrt{11}-3$.

(C) Given equation is $2x^{2} + \frac{5}{3}x - 2 = 0$.
On multiplying both sides by $3$,we get:
$6x^{2} + 5x - 6 = 0$.
To factorise,we split the middle term such that the product of the two numbers is $6 \times (-6) = -36$ and their sum is $5$. The numbers are $9$ and $-4$.
$6x^{2} + 9x - 4x - 6 = 0$
Taking common terms:
$3x(2x + 3) - 2(2x + 3) = 0$
$(2x + 3)(3x - 2) = 0$
Now,setting each factor to zero:
$2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$
$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$
Thus,the roots of the equation are $-\frac{3}{2}$ and $\frac{2}{3}$.

(B) Given equation is $\frac{2}{5} x^{2}-x-\frac{3}{5}=0$.
On multiplying by $5$ on both sides,we get:
$2x^{2}-5x-3=0$
To factorise,we split the middle term $-5x$ into $-6x + x$:
$2x^{2}-6x+x-3=0$
Grouping the terms:
$2x(x-3)+1(x-3)=0$
Taking $(x-3)$ as a common factor:
$(x-3)(2x+1)=0$
Setting each factor to zero:
$x-3=0 \Rightarrow x=3$
$2x+1=0 \Rightarrow x=-\frac{1}{2}$
Hence,the roots of the equation are $-\frac{1}{2}$ and $3$.

(A) Given equation is $3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0$.
To factorise,we split the middle term $-5x$ into $-6x + x$ such that their product is $(3 \sqrt{2}) \times (-\sqrt{2}) = -6$.
$3 \sqrt{2} x^{2}-6 x+x-\sqrt{2}=0$
$3 \sqrt{2} x(x-\sqrt{2})+1(x-\sqrt{2})=0$
$(x-\sqrt{2})(3 \sqrt{2} x+1)=0$
Setting each factor to zero:
$x-\sqrt{2}=0 \Rightarrow x=\sqrt{2}$
$3 \sqrt{2} x+1=0 \Rightarrow x=-\frac{1}{3 \sqrt{2}}$.
Rationalizing the denominator: $x = -\frac{1 \times \sqrt{2}}{3 \sqrt{2} \times \sqrt{2}} = -\frac{\sqrt{2}}{3 \times 2} = -\frac{\sqrt{2}}{6}$.
Thus,the roots are $-\frac{\sqrt{2}}{6}$ and $\sqrt{2}$.

(B) Given equation is $3x^{2} + 5\sqrt{5}x - 10 = 0$.
To factorise,we split the middle term such that the product is $3 \times (-10) = -30$ and the sum is $5\sqrt{5}$.
The factors are $6\sqrt{5}$ and $-\sqrt{5}$,since $(6\sqrt{5}) \times (-\sqrt{5}) = -6 \times 5 = -30$ and $6\sqrt{5} - \sqrt{5} = 5\sqrt{5}$.
$3x^{2} + 6\sqrt{5}x - \sqrt{5}x - 10 = 0$
$3x(x + 2\sqrt{5}) - \sqrt{5}(x + 2\sqrt{5}) = 0$
$(x + 2\sqrt{5})(3x - \sqrt{5}) = 0$
Setting each factor to zero:
$x + 2\sqrt{5} = 0 \Rightarrow x = -2\sqrt{5}$
$3x - \sqrt{5} = 0 \Rightarrow x = \frac{\sqrt{5}}{3}$
Thus,the roots are $-2\sqrt{5}$ and $\frac{\sqrt{5}}{3}$.

(C) Given equation is $21 x^{2}-2 x+\frac{1}{21}=0$.
To simplify,multiply the entire equation by $21$ on both sides:
$21(21 x^{2}-2 x+\frac{1}{21}) = 21(0)$
$441 x^{2}-42 x+1=0$
Now,factorise the quadratic equation by splitting the middle term:
$441 x^{2}-21 x-21 x+1=0$
Group the terms:
$(441 x^{2}-21 x)-(21 x-1)=0$
$21 x(21 x-1)-1(21 x-1)=0$
Factor out $(21 x-1)$:
$(21 x-1)(21 x-1)=0$
Setting each factor to zero:
$21 x-1=0 \Rightarrow x=\frac{1}{21}$
$21 x-1=0 \Rightarrow x=\frac{1}{21}$
Thus,the roots of the equation are $\frac{1}{21}$ and $\frac{1}{21}$.

(A) Given equation: $6x^2 - 7x + 2 = 0$.
Comparing with $ax^2 + bx + c = 0$, we get $a = 6, b = -7, c = 2$.
Discriminant $D = b^2 - 4ac = (-7)^2 - 4(6)(2) = 49 - 48 = 1$.
Since $D > 0$, the equation has two distinct real roots.
To solve by completing the square, divide by $6$: $x^2 - (7/6)x + 1/3 = 0$.
Rewrite as $x^2 - 2(7/12)x = -1/3$.
Add $(7/12)^2$ to both sides: $x^2 - 2(7/12)x + (7/12)^2 = -1/3 + 49/144$.
$(x - 7/12)^2 = (-48 + 49) / 144 = 1/144$.
Taking square root: $x - 7/12 = \pm 1/12$.
Case $1$: $x = 7/12 + 1/12 = 8/12 = 2/3$.
Case $2$: $x = 7/12 - 1/12 = 6/12 = 1/2$.
Thus, the roots are $1/2$ and $2/3$.

(A) Let her actual marks be $x$.
According to the problem,if she scored $10$ more marks,the new score would be $(x + 10)$.
The condition states that $9$ times these marks would be the square of her actual marks:
$9(x + 10) = x^2$
$9x + 90 = x^2$
Rearranging the terms to form a quadratic equation:
$x^2 - 9x - 90 = 0$
Factoring the quadratic equation:
$x^2 - 15x + 6x - 90 = 0$
$x(x - 15) + 6(x - 15) = 0$
$(x + 6)(x - 15) = 0$
This gives $x = -6$ or $x = 15$.
Since marks cannot be negative,we discard $x = -6$.
Therefore,the actual marks obtained by Ajita is $15$.

(B) Let the original average speed of the train be $x \, km/h$.
According to the problem,the time taken for the first part is $\frac{63}{x} \, hours$ and for the second part is $\frac{72}{x+6} \, hours$.
The total time taken is $3 \, hours$,so:
$\frac{63}{x} + \frac{72}{x+6} = 3$
Dividing the entire equation by $9$:
$\frac{7}{x} + \frac{8}{x+6} = \frac{1}{3}$
Multiplying by $3x(x+6)$ to clear the denominators:
$21(x+6) + 24x = x(x+6)$
$21x + 126 + 24x = x^2 + 6x$
$45x + 126 = x^2 + 6x$
$x^2 - 39x - 126 = 0$
Factoring the quadratic equation:
$(x - 42)(x + 3) = 0$
This gives $x = 42$ or $x = -3$.
Since speed cannot be negative,the original average speed is $42 \, km/h$.