Question

The difference between the roots of the quadratic equation x^2−14x+q=0 is 6. Find q.

Answer 1

Answer :

The value of q for, the given quadratic equation is 40

Step-by-step explanation :

Given quadratic equation as :

x² - 14 x + q = 0

And  , Difference between the roots of equation is 6

Let A , B be the roots of the equation

So, A - B = 6

The roots of the quadratic equation  ax² + bx + c = 0 as can be find as :

x = $\frac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$

x = $\frac{14\pm \sqrt{(-14)^{2}-4\times 1\times q}}{2\times 1}$

or, x = $\frac{-14\pm \sqrt{196-4 q}}{2}$

Or, x = $\frac{-14\pm \sqrt{196-4 q}}{2}$

So , The roots are

A = $-7 + \frac{\sqrt{196-4q}}{2}$

And B = $-7 - \frac{\sqrt{196-4q}}{2}$

∵ The difference between the roots is 6

So, A - B = 6

Or, ( $-7 + \frac{\sqrt{196-4q}}{2}$ ) - (  $-7 - \frac{\sqrt{196-4q}}{2}$ ) = 6

Or, ( - 7 + 7 ) + 2 ( $\sqrt{196-4q}$ = 6

Or, 0 + 2 ( $\sqrt{196-4q}$ = 6

∴ 196 - 4 q = 36

or, 4 q = 196 - 36

or 4 q = 160

∴ q = $\frac{160}{4}$

I.e q = 40

S0, The value of q = 40

Hence The value of q for, the given quadratic equation is 40 . Answer