Let p(x) = x^2 - 5x + a and q(x) = x^2 - 3x + | vedclass

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

Question

(D) Given $p(x) = x^2 - 5x + a$ and $q(x) = x^2 - 3x + b$.Since $(x - 1)$ is the $	ext{HCF}$,$p(1) = 0$ and $q(1) = 0$.$p(1) = 1 - 5 + a = 0 \implies

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Given $p(x) = x^2 - 5x + a$ and $q(x) = x^2 - 3x + b$.Since $(x - 1)$ is the $	ext{HCF}$,$p(1) = 0$ and $q(1) = 0$.$p(1) = 1 - 5 + a = 0 \implies a = 4$.$q(1) = 1 - 3 + b = 0 \implies b = 2$.Thus,$p(x) = x^2 - 5x + 4 = (x - 1)(x - 4)$ and $q(x) = x^2 - 3x + 2 = (x - 1)(x - 2)$.$k(x) = 	ext{LCM}(p(x), q(x)) = (x - 1)(x - 2)(x - 4)$.We need the sum of the roots of $(x - 1) + k(x) = 0$.$(x - 1) + (x - 1)(x - 2)(x - 4) = 0$.$(x - 1)[1 + (x - 2)(x - 4)] = 0$.$(x - 1)[1 + x^2 - 6x + 8] = 0$.$(x - 1)(x^2 - 6x + 9) = 0$.$(x - 1)(x - 3)^2 = 0$.The roots are $1, 3, 3$.The sum of the roots is $1 + 3 + 3 = 7$.

Let $p(x) = x^2 - 5x + a$ and $q(x) = x^2 - 3x + b$,where $a$ and $b$ are positive integers. Suppose $\text{HCF}(p(x), q(x)) = x - 1$ and $k(x) = \text{LCM}(p(x), q(x))$. If the coefficient of the highest degree term of $k(x)$ is $1$,then the sum of the roots of $(x - 1) + k(x)$ is:

Similar Questions

If $\alpha$ and $\alpha^2$ are the roots of the equation $x^2 + x + 1 = 0$,then the equation whose roots are $\alpha^{31}$ and $\alpha^{62}$ is .....

If $\sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} = x + 5$,then $x = \dots$

If $a, b, c$ are in Arithmetic Progression $(AP)$,then the roots of the equation $ax^2 - 2bx + c = 0$ are

If $\sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} = x + 5$,then $x$ is equal to

The number of real solutions of the equation $3(x^2 + \frac{1}{x^2}) - 2(x + \frac{1}{x}) + 5 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module