The integral values of a for which the quadra | vedclass

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

Question

(A) The given quadratic equation is $(x - a)(x - 10) + 1 = 0$.Expanding this,we get $x^{2} - (10 + a)x + 10a + 1 = 0$.For the roots to be integers,the

Vedclass · Accepted Answer

$8, 12$

Vedclass · Answer

(A) The given quadratic equation is $(x - a)(x - 10) + 1 = 0$.Expanding this,we get $x^{2} - (10 + a)x + 10a + 1 = 0$.For the roots to be integers,the discriminant $D$ must be a perfect square.$D = (10 + a)^{2} - 4(10a + 1) = 100 + 20a + a^{2} - 40a - 4 = a^{2} - 20a + 96$.We can rewrite this as $D = (a - 10)^{2} - 4$.Let $D = k^{2}$ for some non-negative integer $k$.Then $(a - 10)^{2} - k^{2} = 4$,which implies $(a - 10 - k)(a - 10 + k) = 4$.Since $a$ and $k$ are integers,the factors must be pairs of integers whose product is $4$.The possible cases are:$1) (a - 10 - k) = 2$ and $(a - 10 + k) = 2 \Rightarrow a - 10 = 2 \Rightarrow a = 12$.$2) (a - 10 - k) = -2$ and $(a - 10 + k) = -2 \Rightarrow a - 10 = -2 \Rightarrow a = 8$.$3) (a - 10 - k) = 1$ and $(a - 10 + k) = 4$ (No integer solution for $a$).Thus,the integral values of $a$ are $8$ and $12$.

The integral values of $a$ for which the quadratic equation $(x - a)(x - 10) + 1 = 0$ has integral roots are

Similar Questions

If $2+4i$ is one of the roots of $x^2+bx+c=0$ with $b, c \in R$,then $(b, c) =$

The number of distinct real solutions for the equation $|x^2+2x-8|+x-2=0$ is

The number of real solutions of the equation $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$ is:

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation $6x^2 - 11x + \alpha = 0$ are rational numbers is:

The roots of the equations $2x^2 - 5x + 1 = 0$ and $x^2 + 5x + 2 = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module