Question

If $x_1,x_2$ are the roots of the equation $5x^2+bx-28=0\&5x_1+2x_2=1\forall b\in Z.$ Then find the integral values of $b.$

• A. $-13$
• B. $13$
• C. None of the above.
• D. $-31$

Answer 1

The correct option is A $-13$

Given $x_1,x_2$ are the roots of the quadratic equation $5x^2+bx-28=0$

By the relation of sum and product of roots, we get:
$x_1+x_2=-\frac{b}{5}...\left(i\right)$

$x_1.x_2=\frac{-28}{5}\cdots \left(ii\right)$

$\text{Given}5x_1+2x_2=1\cdots \left(iii\right)$

From $\left(iii\right),x_1=\frac{1-2x_2}{5}\cdots \left(iv\right)$

$\Rightarrow (\frac{1-2x_2}{5})x_2=-\frac{28}{5}$

$\Rightarrow 2x_2^2-x_2-28=0$

$\Rightarrow 2x_2^2-8x_2+7x_2-28=0$

$\Rightarrow 2x_2(x_2-4)+7(x_2-4)=0$

$\Rightarrow (2x_2+7)(x_2-4)=0$

$\Rightarrow x_2=\frac{-7}{2},4$

On substituting these values in the equation $\left(iv\right)$
At $x_2=\frac{-7}{2}$ $x_1=\frac{1-2\times \frac{-7}{2}}{5}=\frac{8}{5}$

And at $x_2=4$

$x_1=\frac{1-2\left(4\right)}{5}=\frac{-7}{5}$

If we take $x_1=\frac{8}{5}$ and $x_2=\frac{-7}{2}$
then from $\left(i\right)$ we get $b=\frac{19}{2}.$ This value is not an integer.

If we take the second case in which $x_1=\frac{-7}{5}$ and $x_2=4$, then from $\left(i\right)$ we get $b=-13$
This is the required integer value.