If the polynomial P(x) = x^2 + ax + b has fac | vedclass

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

Question

(D) Given $P(x) = x^2 + ax + b = (x - a)(x - b)$.Expanding the right side: $x^2 - (a + b)x + ab = x^2 + ax + b$.Comparing coefficients:$a = -(a + b)$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given $P(x) = x^2 + ax + b = (x - a)(x - b)$.Expanding the right side: $x^2 - (a + b)x + ab = x^2 + ax + b$.Comparing coefficients:$a = -(a + b)$ $\Rightarrow 2a + b = 0$ $\Rightarrow b = -2a$.$b = ab$.Case $1$: If $b 
eq 0$,then $a = 1$. Substituting into $b = -2a$,we get $b = -2$.Then $P(x) = x^2 + x - 2$. $P(2) = 2^2 + 2 - 2 = 4$.Case $2$: If $b = 0$,then $2a + 0 = 0 \Rightarrow a = 0$.Then $P(x) = x^2$. $P(2) = 2^2 = 4$.In both cases,$P(2) = 4$.

If the polynomial $P(x) = x^2 + ax + b$ has factors $(x - a)(x - b)$,where $a, b \in R$,then the value of $P(2)$ is:

Similar Questions

If $\alpha$ and $\beta$ are the real roots of the equation $x^2+ax+b=0$,where $\alpha+\beta=\frac{1}{2}$ and $\alpha^3+\beta^3=\frac{37}{8}$,then find the value of $a-\frac{1}{b}$.

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

If the sum of the roots of the equation $ax^2 + bx + c = 0$ is equal to the sum of their squares,then:

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

Let $p$ and $q$ be two positive numbers such that $p + q = 2$ and $p^{4} + q^{4} = 272$. Then $p$ and $q$ are roots of the equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module