Nature of Roots Questions in English

(C) For a quadratic equation of the form $ax^2 + bx + c = 0$,the discriminant $D$ is given by the formula $D = b^2 - 4ac$.
Comparing the given equation $3x^2 - 6x + 4 = 0$ with the standard form,we have $a = 3$,$b = -6$,and $c = 4$.
Substituting these values into the formula:
$D = (-6)^2 - 4(3)(4)$
$D = 36 - 48$
$D = -12$
Therefore,the discriminant is $-12$.

(D) The given quadratic equation is $x^{2} + 12x + 36 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 1$,$b = 12$,and $c = 36$.
The discriminant $D$ is calculated as $D = b^{2} - 4ac$.
$D = (12)^{2} - 4(1)(36) = 144 - 144 = 0$.
Since the discriminant $D = 0$,the roots of the quadratic equation are real and equal.
Alternatively,the equation can be written as $(x + 6)^{2} = 0$,which gives $x = -6, -6$.

(A) To determine the nature of the roots of the quadratic equation $ax^2 + bx + c = 0$,we calculate the discriminant $D = b^2 - 4ac$.
Here,$a = 9$,$b = 30$,and $c = 25$.
Substituting these values into the discriminant formula:
$D = (30)^2 - 4(9)(25)$
$D = 900 - 900$
$D = 0$
Since the discriminant $D = 0$,the quadratic equation has real and equal roots.

(B) The given quadratic equation is $x^{2}-10x+(2k-1)=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1$,$b=-10$,and $c=2k-1$.
The discriminant $D$ is given by the formula $D=b^{2}-4ac$.
Given $D=40$,we substitute the values:
$40 = (-10)^{2} - 4(1)(2k-1)$
$40 = 100 - 4(2k-1)$
$40 = 100 - 8k + 4$
$40 = 104 - 8k$
$8k = 104 - 40$
$8k = 64$
$k = 8$.

(D) For a quadratic equation $ax^{2} + bx + c = 0$,the product of the roots is given by $\alpha \cdot \beta = c/a$.
Given the equation $2x^{2} + 5x + 3 = 0$,we have $a = 2$,$b = 5$,and $c = 3$.
Let the roots be $\alpha = -1$ and $\beta$ be the other root.
Using the product of roots formula: $(-1) \cdot \beta = 3/2$.
Therefore,$\beta = -3/2$.
Alternatively,using the sum of roots: $\alpha + \beta = -b/a$.
$-1 + \beta = -5/2$.
$\beta = -5/2 + 1 = -3/2$.

(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 $x^{2}-6x+9=0$.

(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) 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 $4$.

(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 $-6$.

(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 $(D < 0)$.

(B) The given quadratic equation is $25x^{2} - 10x + 1 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 25$,$b = -10$,and $c = 1$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values:
$D = (-10)^{2} - 4(25)(1)$
$D = 100 - 100$
$D = 0$.
Thus,the value of the discriminant is $0$.

(D) The given quadratic equation is $x^{2}+18x+81=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we have $a=1$,$b=18$,and $c=81$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
Substituting the values: $D = (18)^{2} - 4(1)(81) = 324 - 324 = 0$.
Since the discriminant $D=0$,the quadratic equation has two real and equal roots.
Alternatively,we can factorize the equation: $x^{2}+18x+81 = (x+9)^{2} = 0$.
This gives $(x+9)(x+9)=0$,so $x = -9, -9$.
Thus,the roots are equal.

(C) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Comparing the given equation $x^{2} + 5x + 1 = 0$ with the standard form,we get $a = 1$,$b = 5$,and $c = 1$.
Substituting these values into the formula:
$D = (5)^{2} - 4(1)(1)$
$D = 25 - 4$
$D = 21$.
Therefore,the value of the discriminant is $21$.

(C) The given quadratic equation is $2x^{2} + 5x - k = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 2$,$b = 5$,and $c = -k$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given that $D = 81$,we substitute the values:
$81 = (5)^{2} - 4(2)(-k)$
$81 = 25 + 8k$
$81 - 25 = 8k$
$56 = 8k$
$k = \frac{56}{8} = 7$.
Therefore,the value of $k$ is $7$.

(C) 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.
Given that $D = 0$,the formula becomes $x = \frac{-b \pm \sqrt{0}}{2a}$.
This simplifies to $x = \frac{-b \pm 0}{2a}$.
Therefore,both roots are equal to $x = \frac{-b}{2a}$.

(C) The quadratic equation is given by $ax^{2} + bx + c = 0$.
Given that $b = 0$,the equation simplifies to $ax^{2} + c = 0$.
This implies $ax^{2} = -c$,or $x^{2} = -c/a$.
Taking the square root of both sides,we get $x = \pm \sqrt{-c/a}$.
Let the roots be $\alpha$ and $\beta$. Then $\alpha = \sqrt{-c/a}$ and $\beta = -\sqrt{-c/a}$.
Since $\alpha = -\beta$,the roots are equal in magnitude but opposite in sign.
Therefore,the roots are opposite.

(C) Let the roots of the quadratic equation $25x^{2} - x(m - 2) - 1 = 0$ be $\alpha$ and $-\alpha$ because they are opposite to each other.
For a quadratic equation $ax^{2} + bx + c = 0$,the sum of roots is given by $\alpha + \beta = -b/a$.
Here,$a = 25$,$b = -(m - 2)$,and $c = -1$.
Sum of roots: $\alpha + (-\alpha) = -[-(m - 2)] / 25$.
$0 = (m - 2) / 25$.
Multiplying both sides by $25$,we get $0 = m - 2$.
Therefore,$m = 2$.

(B) The quadratic formula for the roots of the equation $ax^2 + bx + c = 0$ is given by $x = \frac{-b \pm \sqrt{D}}{2a}$,where $D = b^2 - 4ac$ is the discriminant.
Given that the discriminant $D = 0$,we substitute this into the quadratic formula:
$x = \frac{-b \pm \sqrt{0}}{2a}$
$x = \frac{-b \pm 0}{2a}$
$x = -\frac{b}{2a}$
Therefore,when $D = 0$,the quadratic equation has two equal real roots,each equal to $-\frac{b}{2a}$.

(D) quadratic polynomial $p(x) = ax^2 + bx + c$ has no real zeros if its discriminant $D = b^2 - 4ac < 0$.
For option $A$: $D = (-5)^2 - 4(1)(4) = 25 - 16 = 9 > 0$. It has two real zeros.
For option $B$: $D = (2)^2 - 4(1)(1) = 4 - 4 = 0$. It has one real zero.
For option $C$: $D = (0)^2 - 4(1)(-4) = 16 > 0$. It has two real zeros.
For option $D$: $D = (2)^2 - 4(1)(7) = 4 - 28 = -24 < 0$. Since $D < 0$,this polynomial has no real zeros.

(C) For a quadratic equation $ax^{2} + bx + c = 0$,the nature of the roots is determined by the discriminant $D = b^{2} - 4ac$.
Here,$a = 4$,$b = 4$,and $c = -3$.
Calculating the discriminant: $D = (4)^{2} - 4(4)(-3) = 16 + 48 = 64$.
Since $D > 0$,the roots are real and unequal.
Since $D = 64$ is a perfect square,the roots are rational.
Therefore,the roots are unequal and rational.

(A) The discriminant $D$ of a quadratic equation $ax^{2}+bx+c=0$ is given by $D = b^{2}-4ac$.
$1.$ For $x^{2}+5x+6=0$,$D = 5^{2}-4(1)(6) = 25-24 = 1$. Thus,$(1-a)$.
$2.$ For $x^{2}+5x+4=0$,$D = 5^{2}-4(1)(4) = 25-16 = 9$. Thus,$(2-b)$.
$3.$ For $x^{2}+4x+3=0$,$D = 4^{2}-4(1)(3) = 16-12 = 4$. Thus,$(3-c)$.
$4.$ For $x^{2}+6x+5=0$,$D = 6^{2}-4(1)(5) = 36-20 = 16$. Thus,$(4-d)$.
Matching these,we get $(1-a), (2-b), (3-c), (4-d)$. Note: The provided options in the prompt were inconsistent with the calculation; based on the standard calculation,the correct match is $(1-a), (2-b), (3-c), (4-d)$.