Mix Examples - Quadratic Equations Questions in English

(A) quadratic equation in one variable is an equation of the form $ax^{2} + bx + c = 0$,where $a, b, c$ are real numbers and $a \neq 0$.
Option $A$: $3x^{2} + \sqrt{2}x - 7 = 0$ is a quadratic equation in one variable $(x)$.
Option $B$: $3x^{2} + 2y - 6 = 0$ has two variables ($x$ and $y$).
Option $C$: $4x - y = 5$ is a linear equation in two variables.
Option $D$: $5x^{3} - 1 = 2$ is a cubic equation in one variable.

(C) quadratic equation is an equation of the form $ax^2 + bx + c = 0$,where $a \neq 0$.
Let us simplify each option:
$A$: $x(3x + 7) = (x + 1)(x - 1) \implies 3x^2 + 7x = x^2 - 1 \implies 2x^2 + 7x + 1 = 0$. This is a quadratic equation.
$B$: $x^2 - 2x + 1 = 0$. This is already in the form $ax^2 + bx + c = 0$ with $a = 1$. This is a quadratic equation.
$C$: $2x(3x - 5) + 1 = 3x(2x + 5) + 3 \implies 6x^2 - 10x + 1 = 6x^2 + 15x + 3 \implies -10x + 1 = 15x + 3 \implies 25x + 2 = 0$. This is a linear equation,not a quadratic equation.
$D$: $4 - 3x - 2x^2 = 0$. This is a quadratic equation with $a = -2$.

(B) To determine the nature of the given expression,we expand the product:
$(x+6)(x+5) = x^2 + 5x + 6x + 30 = x^2 + 11x + 30$.
Since the equation is given as $(x+6)(x+5) = 0$,we can write it as $x^2 + 11x + 30 = 0$.
An equation of the form $ax^2 + bx + c = 0$,where $a \neq 0$,is defined as a quadratic equation.
Since the highest power of the variable $x$ is $2$,it is a quadratic equation.

(A) quadratic equation is an equation of the form $ax^2 + bx + c = 0$,where $a, b, c$ are real numbers and $a \neq 0$.
Let us analyze each option:
$A$: $x + \frac{1}{x} = 2 \implies x^2 + 1 = 2x \implies x^2 - 2x + 1 = 0$. This is a quadratic equation.
$B$: $x^2 + \frac{1}{x} = 2 \implies x^3 + 1 = 2x \implies x^3 - 2x + 1 = 0$. This is a cubic equation.
$C$: $x + \frac{1}{x^2} = 3 \implies x^3 + 1 = 3x^2 \implies x^3 - 3x^2 + 1 = 0$. This is a cubic equation.
$D$: $x(x + 1) = 2 \implies x^2 + x = 2 \implies x^2 + x - 2 = 0$. This is also a quadratic equation.
Note: In the original provided options,$A$ is the standard form representation. If multiple options are valid,$A$ is the most direct representation.

(B) Given the quadratic equation: $x^{2} + 5x + 6 = 0$.
To solve this,we factorize the quadratic expression by splitting the middle term:
$x^{2} + 2x + 3x + 6 = 0$
$x(x + 2) + 3(x + 2) = 0$
$(x + 2)(x + 3) = 0$
Setting each factor to zero:
$x + 2 = 0 \implies x = -2$
$x + 3 = 0 \implies x = -3$
Therefore,the solution set is $\{-2, -3\}$.

(A) To find the roots of the quadratic equation $x^{2}-2x-15=0$,we can use the factorization method.
We need to find two numbers whose product is $-15$ and whose sum is $-2$.
These two numbers are $-5$ and $3$,since $(-5) \times 3 = -15$ and $(-5) + 3 = -2$.
Now,rewrite the equation as: $x^{2}-5x+3x-15=0$.
Factor by grouping: $x(x-5)+3(x-5)=0$.
This gives: $(x-5)(x+3)=0$.
Setting each factor to zero,we get $x-5=0$ or $x+3=0$.
Therefore,$x=5$ or $x=-3$.
The roots of the equation are $5$ and $-3$.

(D) To find the solution set of the quadratic equation $x^{2}+5x-14=0$,we factorize the quadratic expression.
We look for two numbers whose product is $-14$ and whose sum is $5$.
These numbers are $7$ and $-2$,since $7 \times (-2) = -14$ and $7 + (-2) = 5$.
Thus,we can write the equation as: $x^{2} + 7x - 2x - 14 = 0$.
Grouping the terms,we get: $x(x + 7) - 2(x + 7) = 0$.
Factoring out $(x + 7)$,we have: $(x + 7)(x - 2) = 0$.
Setting each factor to zero,we get $x + 7 = 0$ or $x - 2 = 0$.
Therefore,$x = -7$ or $x = 2$.
The solution set is $\{-7, 2\}$.

(A) To determine which equation has $x=2$ as a solution,we substitute $x=2$ into each given equation and check if the result is $0$.
For option $A$: $x^{2}-3x+2=0$
Substitute $x=2$: $(2)^{2}-3(2)+2 = 4-6+2 = 0$.
Since the result is $0$,$x=2$ is a solution for this equation.
For option $B$: $x^{2}+x+2=0$
Substitute $x=2$: $(2)^{2}+2+2 = 4+2+2 = 8 \neq 0$.
For option $C$: $x^{2}-2x+2=0$
Substitute $x=2$: $(2)^{2}-2(2)+2 = 4-4+2 = 2 \neq 0$.
For option $D$: $2x^{2}+2x+1=0$
Substitute $x=2$: $2(2)^{2}+2(2)+1 = 2(4)+4+1 = 8+4+1 = 13 \neq 0$.
Therefore,the correct option is $A$.

(B) Given that $3$ is a root of the quadratic equation $x^{2}-kx+6=0$.
Substituting $x=3$ into the equation:
$(3)^{2}-k(3)+6=0$
$9-3k+6=0$
$15-3k=0$
$3k=15$
$k=5$
Therefore,the value of $k$ is $5$.

(B) Since $-3$ is a root of the quadratic equation $x^{2}+3(k+2)x-9=0$,it must satisfy the equation.
Substituting $x=-3$ into the equation:
$(-3)^{2}+3(k+2)(-3)-9=0$
$9-9(k+2)-9=0$
$9-9k-18-9=0$
$-9k-18=0$
$-9k=18$
$k=-2$

(B) Given equation: $(x-7)^{2}-16=0$
Method $1$: Using the identity $a^2 - b^2 = (a-b)(a+b)$
$(x-7)^2 - 4^2 = 0$
$(x-7-4)(x-7+4) = 0$
$(x-11)(x-3) = 0$
Therefore,$x-11=0$ or $x-3=0$,which gives $x=11$ or $x=3$.
Method $2$: Expanding the square
$x^2 - 14x + 49 - 16 = 0$
$x^2 - 14x + 33 = 0$
$x^2 - 11x - 3x + 33 = 0$
$x(x-11) - 3(x-11) = 0$
$(x-11)(x-3) = 0$
Thus,the roots are $3$ and $11$.

(B) Given the quadratic equation: $x(x+1)-6=0$
Expanding the equation: $x^{2}+x-6=0$
To find the roots,we factorize the quadratic expression:
$x^{2}+3x-2x-6=0$
$x(x+3)-2(x+3)=0$
$(x+3)(x-2)=0$
Setting each factor to zero:
$x+3=0 \implies x=-3$
$x-2=0 \implies x=2$
Therefore,the roots are $-3$ and $2$.

(A) Given that $1$ is the solution (root) of the quadratic equation $3x^{2}-kx+2=0$.
Substituting $x=1$ into the equation:
$3(1)^{2}-k(1)+2=0$
$3(1)-k+2=0$
$3-k+2=0$
$5-k=0$
Therefore,$k=5$.

(B) Given that $x=-2$ is a solution of the quadratic equation $k x^{2}+5 x+2=0$.
Substituting $x=-2$ into the equation:
$k(-2)^{2}+5(-2)+2=0$
$k(4)-10+2=0$
$4k-8=0$
$4k=8$
$k=2$
Therefore,the value of $k$ is $2$.

(B) Given that $-3$ is a solution of the quadratic equation $2x^{2} + 5x + k = 0$.
Substituting $x = -3$ into the equation:
$2(-3)^{2} + 5(-3) + k = 0$
$2(9) - 15 + k = 0$
$18 - 15 + k = 0$
$3 + k = 0$
$k = -3$

(C) Given quadratic equation: $6 x^{2}-13 x+6=0$
Splitting the middle term: $6 x^{2}-9 x-4 x+6=0$
Factoring by grouping: $3 x(2 x-3)-2(2 x-3)=0$
$(2 x-3)(3 x-2)=0$
Setting each factor to zero: $2 x-3=0$ or $3 x-2=0$
Therefore,the roots are $x=\frac{3}{2}$ or $x=\frac{2}{3}$.

(A) Given equation: $x + \frac{2}{x} = 3$
Multiply the entire equation by $x$ to eliminate the denominator:
$x^2 + 2 = 3x$
Rearrange the equation into the standard form $ax^2 + bx + c = 0$:
$x^2 - 3x + 2 = 0$
Factor the quadratic equation:
$x^2 - 2x - x + 2 = 0$
$x(x - 2) - 1(x - 2) = 0$
$(x - 2)(x - 1) = 0$
Setting each factor to zero:
$x - 2 = 0 \implies x = 2$
$x - 1 = 0 \implies x = 1$
Thus,the roots are $2$ and $1$.

(B) quadratic equation has recurring (equal) roots if its discriminant $D = b^{2}-4ac$ is equal to $0$.
For option $A$: $x^{2}-6x-8=0$,$D = (-6)^{2}-4(1)(-8) = 36+32 = 68 \neq 0$.
For option $B$: $x^{2}-6x+9=0$,$D = (-6)^{2}-4(1)(9) = 36-36 = 0$. Since $D=0$,the roots are recurring (equal).
For option $C$: $x^{2}-9=0$,$D = (0)^{2}-4(1)(-9) = 36 \neq 0$.
For option $D$: $x^{2}-4=0$,$D = (0)^{2}-4(1)(-4) = 16 \neq 0$.
Therefore,the equation with recurring roots is $x^{2}-6x+9=0$.

(D) quadratic expression $ax^2 + bx + c$ is a perfect square if its discriminant $D = b^2 - 4ac$ is equal to $0$.
Here,$a = 16$,$b = 40$,and $c = k$.
Setting the discriminant to zero: $D = (40)^2 - 4(16)(k) = 0$.
$1600 - 64k = 0$.
$64k = 1600$.
$k = \frac{1600}{64} = 25$.
Alternatively,for a perfect square of the form $(mx + n)^2 = m^2x^2 + 2mnx + n^2$,we compare $16x^2 + 40x + k$ with $(4x + n)^2 = 16x^2 + 8nx + n^2$.
Equating the middle terms: $8n = 40$,so $n = 5$.
Then $k = n^2 = 5^2 = 25$.

(B) The golden ratio,denoted by $\phi$,is defined as the ratio of two quantities such that the ratio of the sum of the quantities to the larger quantity is equal to the ratio of the larger quantity to the smaller one.
Mathematically,for two quantities $a$ and $b$ where $a > b > 0$,the golden ratio is defined as $\frac{a+b}{a} = \frac{a}{b} = \phi$.
This leads to the quadratic equation $\phi^2 - \phi - 1 = 0$.
Using the quadratic formula $\phi = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $\phi = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}$.
Since the ratio must be positive,we take the positive root: $\phi = \frac{1+\sqrt{5}}{2} \approx 1.618$.

(C) Let $x = \frac{1+\sqrt{5}}{2}$.
Multiply both sides by $2$: $2x = 1+\sqrt{5}$.
Subtract $1$ from both sides: $2x-1 = \sqrt{5}$.
Square both sides: $(2x-1)^{2} = (\sqrt{5})^{2}$.
Expand the left side: $4x^{2}-4x+1 = 5$.
Rearrange the terms: $4x^{2}-4x-4 = 0$.
Divide the entire equation by $4$: $x^{2}-x-1 = 0$.
Thus,the golden number is a root of the quadratic equation $x^{2}-x-1=0$.

(C) Given the quadratic equation $x^{2} = 49$.
Subtracting $49$ from both sides,we get $x^{2} - 49 = 0$.
This is in the form of $a^{2} - b^{2} = (a - b)(a + b)$,where $a = x$ and $b = 7$.
So,$(x - 7)(x + 7) = 0$.
Setting each factor to zero,we get $x - 7 = 0$ or $x + 7 = 0$.
Therefore,$x = 7$ or $x = -7$.

(D) Given equation: $(x+2)^{2} = 16$
Taking the square root on both sides: $x+2 = \pm 4$
Case $1$: $x+2 = 4 \implies x = 4-2 = 2$
Case $2$: $x+2 = -4 \implies x = -4-2 = -6$
Therefore,the roots are $2$ and $-6$.

(D) quadratic equation with roots $\alpha$ and $\beta$ is given by the formula: $(x - \alpha)(x - \beta) = 0$.
Given the roots are $\alpha = -1$ and $\beta = 2$.
Substituting these values into the formula,we get: $(x - (-1))(x - 2) = 0$.
This simplifies to: $(x + 1)(x - 2) = 0$.
Expanding the expression: $x(x) + x(-2) + 1(x) + 1(-2) = 0$.
$x^{2} - 2x + x - 2 = 0$.
$x^{2} - x - 2 = 0$.
Therefore,the correct quadratic equation is $x^{2} - x - 2 = 0$.

(D) quadratic equation with roots $\alpha$ and $\beta$ is given by the formula: $x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta) = 0$.
Here,the roots are $\alpha = -8$ and $\beta = 8$.
Sum of the roots: $\alpha + \beta = -8 + 8 = 0$.
Product of the roots: $\alpha \cdot \beta = (-8) \times (8) = -64$.
Substituting these values into the formula:
$x^{2} - (0)x + (-64) = 0$
$x^{2} - 64 = 0$
$x^{2} = 64$.
Thus,the correct option is $D$.

(C) For a quadratic equation in the standard form $ax^2 + bx + c = 0$,where $a \neq 0$,the discriminant is defined as $D = b^2 - 4ac$.
This value determines the nature of the roots of the quadratic equation.
The symbol used to represent the discriminant is $D$.

(C) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant is given by $D = b^2 - 4ac$.
$1$. If $D > 0$,the equation has two distinct real roots.
$2$. If $D = 0$,the equation has two equal real roots.
$3$. If $D < 0$,the square root of $D$ is not a real number,therefore the quadratic equation has no real roots.
Thus,the condition for the non-existence of real roots is $D < 0$.

(B) For a quadratic equation $ax^2 + bx + c = 0$ with rational coefficients $a, b, c \in \mathbb{Q}$,the roots are given by the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a}$,where $D = b^2 - 4ac$ is the discriminant.
$1$. If $D > 0$ and $D$ is a perfect square of a rational number,then $\sqrt{D}$ is rational,making the roots rational and distinct.
$2$. If $D = 0$,the roots are real and equal.
$3$. If $D < 0$,the roots are complex (imaginary).
Therefore,for the roots to be distinct and rational,$D$ must be greater than $0$ and a perfect square of a rational number,provided $a, b, c$ are rational.

(D) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant is given by $D = b^2 - 4ac$.
If $D > 0$,the equation has two distinct real roots.
If $D = 0$,the equation has two equal real roots.
If $D < 0$,the equation has no real roots (complex roots).
Therefore,for the roots to be distinct and real,the condition is $D > 0$.

(D) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
If $D > 0$,the equation has two distinct real roots.
If $D = 0$,the equation has two equal real roots.
If $D < 0$,the equation has no real roots.
Therefore,for the roots to be equal,the condition is $D = 0$.

(A) For a quadratic equation $ax^2 + bx + c = 0$,the roots are given by the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a}$,where $D = b^2 - 4ac$ is the discriminant.
This formula provides two roots: $x_1 = \frac{-b + \sqrt{D}}{2a}$ and $x_2 = \frac{-b - \sqrt{D}}{2a}$.
Since the question states that one root is $\frac{-b + \sqrt{D}}{2a}$,the other root must be $\frac{-b - \sqrt{D}}{2a}$.

(C) The standard form of a quadratic equation is given by $ax^{2}+bx+c=0$.
Given the equation $5x^{2}-4x-1=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we identify the coefficients:
$a = 5$
$b = -4$
$c = -1$
Therefore,the value of $b$ is $-4$.

(C) The given quadratic equation is $x(2x - 1) - 5 = 0$.
Expanding the equation,we get $2x^2 - x - 5 = 0$.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Comparing $2x^2 - x - 5 = 0$ with $ax^2 + bx + c = 0$,we identify the coefficients as $a = 2$,$b = -1$,and $c = -5$.
Therefore,the value of $a$ is $2$.

(C) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
If the roots of the quadratic equation are equal,then the discriminant must be equal to zero,i.e.,$D = 0$.
Substituting the value of $D$,we get $b^2 - 4ac = 0$.
Therefore,$b^2 = 4ac$.

(C) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given the equation $4x^{2} + 20x + 25 = 0$,we identify the coefficients as $a = 4$,$b = 20$,and $c = 25$.
Substituting these values into the formula:
$D = (20)^{2} - 4(4)(25)$
$D = 400 - 400$
$D = 0$
Thus,the value of the discriminant is $0$.

(C) For a quadratic equation in the form $ax^{2} + bx + c = 0$,the discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given the equation $5x^{2} + x + 2 = 0$,we identify the coefficients as $a = 5$,$b = 1$,and $c = 2$.
Substituting these values into the formula:
$D = (1)^{2} - 4(5)(2)$
$D = 1 - 40$
$D = -39$
Therefore,the value of the discriminant is $-39$.

(D) For the quadratic equation $ax^{2}+bx+c=0$,the discriminant $D$ is given by the formula $D=b^{2}-4ac$.
Given the equation $x^{2}-8x+15=0$,we identify the coefficients as $a=1, b=-8$,and $c=15$.
Substituting these values into the formula:
$D=(-8)^{2}-4(1)(15)$
$D=64-60$
$D=4$
Therefore,the value of the discriminant is $4$.

(C) The given quadratic equation is $kx^{2} - 4x - 4 = 0$. Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = k$,$b = -4$,and $c = -4$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given that $D = 64$,we substitute the values into the formula:
$(-4)^{2} - 4(k)(-4) = 64$
$16 + 16k = 64$
$16k = 64 - 16$
$16k = 48$
$k = \frac{48}{16}$
$k = 3$

(A) For a quadratic equation of the form $ax^{2}+bx+c=0$,the discriminant $D$ is given by $D = b^{2}-4ac$.
Given equation: $6x^{2}-x-k=0$.
Here,$a=6, b=-1, c=-k$.
Given that the discriminant $D = 25$.
Substituting the values into the formula: $(-1)^{2}-4(6)(-k) = 25$.
$1+24k = 25$.
$24k = 25-1$.
$24k = 24$.
$k = 1$.

(C) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
Given the equation $kx^2 - 6x + 1 = 0$,we have $a = k$,$b = -6$,and $c = 1$.
Since the discriminant is $0$,we set $D = 0$:
$(-6)^2 - 4(k)(1) = 0$
$36 - 4k = 0$
$4k = 36$
$k = 9$.

(B) For a quadratic equation $ax^{2} + bx + c = 0$,the roots are equal if and only if the discriminant $D = b^{2} - 4ac = 0$.
Here,$a = k$,$b = 4$,and $c = 1$.
Substituting these values into the discriminant formula:
$D = (4)^{2} - 4(k)(1) = 0$
$16 - 4k = 0$
$4k = 16$
$k = 4$
Therefore,the correct value of $k$ is $4$.

(D) Since $-4$ is a root of the quadratic equation $x^{2}-ax-8=0$,it must satisfy the equation.
Substituting $x = -4$ into the equation:
$(-4)^{2} - a(-4) - 8 = 0$
$16 + 4a - 8 = 0$
$8 + 4a = 0$
$4a = -8$
$a = -2$

(A) For a quadratic equation of the form $ax^{2}+bx+c=0$,the sum of the roots is given by $\alpha+\beta = -\frac{b}{a}$.
Here,$a=1$,$b=4$,and $c=m$.
Given that one root $\alpha = 2$,let the other root be $\beta$.
Substituting the values into the sum formula: $2 + \beta = -\frac{4}{1}$.
$2 + \beta = -4$.
$\beta = -4 - 2$.
$\beta = -6$.
Therefore,the other root is $-6$.

(B) Let the quadratic equation be $x^{2}+bx-15=0$.
Given that one root is $\alpha = 3$.
Let the other root be $\beta$.
For a quadratic equation $ax^{2}+bx+c=0$,the product of the roots is given by $\alpha \cdot \beta = \frac{c}{a}$.
Here,$a = 1$,$b = b$,and $c = -15$.
Substituting the values,we get $3 \cdot \beta = \frac{-15}{1}$.
$3 \cdot \beta = -15$.
$\beta = \frac{-15}{3} = -5$.
Therefore,the other root is $-5$.

(D) quadratic equation with roots $\alpha$ and $\beta$ is given by the formula: $x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta) = 0$.
Given roots are $\alpha = -4$ and $\beta = 5$.
Sum of roots: $\alpha + \beta = -4 + 5 = 1$.
Product of roots: $\alpha \cdot \beta = (-4) \times 5 = -20$.
Substituting these values into the formula: $x^{2} - (1)x + (-20) = 0$.
This simplifies to: $x^{2} - x - 20 = 0$.
Therefore,the correct option is $D$.

(B) The standard form of a quadratic equation is $ax^{2}+bx+c=0$. Comparing $x^{2}-10x+(2k-1)=0$ with the standard form,we get $a=1$,$b=-10$,and $c=2k-1$.
The discriminant $D$ is given by the formula $D = b^{2}-4ac$.
Given that $D = 40$,we substitute the values into the formula:
$(-10)^{2} - 4(1)(2k-1) = 40$
$100 - 4(2k-1) = 40$
Subtract $100$ from both sides:
$-4(2k-1) = 40 - 100$
$-4(2k-1) = -60$
Divide by $-4$:
$2k-1 = 15$
Add $1$ to both sides:
$2k = 16$
Divide by $2$:
$k = 8$.

(B) For a quadratic equation of the form $ax^{2} + bx + c = 0$, the discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given the equation $x^{2} + \sqrt{2}x + 2 = 0$, we identify the coefficients as $a = 1$, $b = \sqrt{2}$, and $c = 2$.
Substituting these values into the formula:
$D = (\sqrt{2})^{2} - 4(1)(2)$
$D = 2 - 8$
$D = -6$
Therefore, the value of the discriminant is $-6$.

(B) First,solve the quadratic equation $x^{2}-5x+6=0$.
Factorizing the equation: $(x-2)(x-3)=0$.
Thus,the roots are $x=2$ or $x=3$.
Case $1$: If $x=2$ is a root of $x^{2}+3x+c=0$,then $(2)^{2}+3(2)+c=0$.
$4+6+c=0 \implies 10+c=0 \implies c=-10$.
Case $2$: If $x=3$ is a root of $x^{2}+3x+c=0$,then $(3)^{2}+3(3)+c=0$.
$9+9+c=0 \implies 18+c=0 \implies c=-18$.
Therefore,the possible values for $c$ are $-10$ or $-18$.

(C) For a quadratic equation in the standard form $ax^2 + bx + c = 0$,where $a \neq 0$,the discriminant $D$ is used to determine the nature of the roots.
The formula for the discriminant is given by $D = b^2 - 4ac$.
This value helps in identifying whether the roots are real and distinct $(D > 0)$,real and equal $(D = 0)$,or imaginary $(D < 0)$.

(D) The quadratic formula used to find the roots of a quadratic equation $ax^2 + bx + c = 0$ is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
This formula is historically known as the 'Shridhar Acharya formula' in India,as the mathematician Shridhar Acharya provided the method for solving quadratic equations by completing the square.