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.$D = b^{2} - 4ac = (10 + a)^{2} - 4(10a + 1)$.$D = 100 + 20a + a^{2} - 40a - 4$.$D = a^{2} - 20a + 96$.Completing the square,$D = (a - 10)^{2} - 4$.Let $D = k^{2}$ for some integer $k \ge 0$.Then $k^{2} = (a - 10)^{2} - 4$,which implies $(a - 10)^{2} - k^{2} = 4$.$(a - 10 - k)(a - 10 + k) = 4$.Since the difference between these two factors is $2k$ (an even number),both factors must be even.The possible pairs of factors of $4$ are $(2, 2)$ or $(-2, -2)$.Case $1$: $a - 10 - k = 2$ and $a - 10 + k = 2$.Adding these,$2(a - 10) = 4 \Rightarrow a - 10 = 2 \Rightarrow a = 12$.Case $2$: $a - 10 - k = -2$ and $a - 10 + k = -2$.Adding these,$2(a - 10) = -4 \Rightarrow a - 10 = -2 \Rightarrow a = 8$.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 a root of the equation $x^2 + px + 12 = 0$ is $4$,while the roots of the equation $x^2 + px + q = 0$ are equal,then the value of $q$ will be

Solve the given two equations and select the correct answer from the given options.
$I.$ $5x^{2} + 2x - 3 = 0$
$II.$ $2y^{2} + 7y + 6 = 0$

$I. x^{2}+10 x+24=0$
$II. 4 y^{2}-17 y+18=0$

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module