Question

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

Answer 1

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* a* c}}{2* a}`

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

or, x =
`(-14\pm √(196-4 q))/(2)`

Or, x =
`(-14\pm √(196-4 q))/(2)`

So , The roots are

A =
`-7 + (√(196-4q))/(2)`

And B =
`-7 - (√(196-4q))/(2)`

∵ The difference between the roots is 6

So, A - B = 6

Or, (
`-7 + (√(196-4q))/(2)` ) - (
`-7 - (√(196-4q))/(2)` ) = 6

Or, ( - 7 + 7 ) + 2 (
`√(196-4q)` = 6

Or, 0 + 2 (
`√(196-4q)` = 6

∴ 196 - 4 q = 36

or, 4 q = 196 - 36

or 4 q = 160

∴ q =
`(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