Let p(x) be a quadratic polynomial with const | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $p(x) = ax^2 + bx + c$. Given,the constant term $c = 1$. Therefore,$p(x) = ax^2 + bx + 1$. By the Remainder Theorem,$p(1) = 2$ and $p(-1) = 4$

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Let $p(x) = ax^2 + bx + c$. Given,the constant term $c = 1$. Therefore,$p(x) = ax^2 + bx + 1$. By the Remainder Theorem,$p(1) = 2$ and $p(-1) = 4$. Substituting $x=1$: $a(1)^2 + b(1) + 1 = 2 \implies a + b = 1$ (Equation $I$). Substituting $x=-1$: $a(-1)^2 + b(-1) + 1 = 4 \implies a - b = 3$ (Equation $II$). Adding Equation $I$ and Equation $II$: $(a+b) + (a-b) = 1 + 3 \implies 2a = 4 \implies a = 2$. Substituting $a=2$ into Equation $I$: $2 + b = 1 \implies b = -1$. Thus,the polynomial is $p(x) = 2x^2 - x + 1$. The sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is given by $-\frac{b}{a}$. Sum of the roots $= -\frac{-1}{2} = \frac{1}{2}$.

Let $p(x)$ be a quadratic polynomial with constant term $1$. Suppose $p(x)$,when divided by $x-1$,leaves remainder $2$ and when divided by $x+1$,leaves remainder $4$. Then,the sum of the roots of $p(x)=0$ is

Similar Questions

If the sum of any two roots of the equation $x^3+p x^2+q x+r=0$ is zero,then

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the roots of the equation $cx^2 + bx + a = 0$ are

If the harmonic mean of the roots of the equation $\sqrt{2} x^2 - bx + (8 - 2\sqrt{5}) = 0$ is $4$,then the value of $b$ is

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+3x-4=0$,then the value of $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$ is

If the ratio of the roots of the equation $12x^2 - mx + 5 = 0$ is $2 : 3$,then $m = .....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module