Question

If the difference of the roots of the equation x2 + kx + 7 = 0 is 6 then possible values of k are ?

Answer 1

Let X1, X2 be the roots.
Given, X1 - X2 = 6.
From the equation, X1 + X2 = -k & X1 × X2 = 7.
Using the identity,
(X1 - X2)^2 = (X1 + X2)^2 - 4X1X2
(6)^2 = (-k)^2 - 4×7
k = 8 or -8

Answer 2

The possible values of k are ± 8.

Step-by-step explanation:

Given: Quadratic equation $x^{2} + kx + 7 = 0$

Difference between the roots = 6

To Find: Value of 'k'

Solution:

• Finding the possible values of k

We have the quadratic equation $x^{2} + kx + 7 = 0$ such that the roots can be found by using the quadratic formula, which is,

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac} }{2a} \ \text{for the quadratic equation } ax^{2} +bx+c = 0$

Therefore, for the quadratic equation $x^{2} + kx + 7 = 0$,

$x = \dfrac{-k \pm \sqrt{k^2 - 28} }{2}$

$\Rightarrow x_1 = \dfrac{-k - \sqrt{k^2 - 28} }{2} \ \& \ x_2 = \dfrac{-k + \sqrt{k^2 - 28} }{2}$

Given the difference between the roots is 6, we can write,

$\Rightarrow x_2 - x_1 = \dfrac{-k + \sqrt{k^2 - 28} }{2} - \dfrac{-k - \sqrt{k^2 - 28} }{2} = 6$

$\Rightarrow x_2 - x_1 = \dfrac{-k + \sqrt{k^2 - 28} }{2} + \dfrac{k + \sqrt{k^2 - 28} }{2} = 6$

$\Rightarrow x_2 - x_1 = \dfrac{-k + \sqrt{k^2 - 28} + k + \sqrt{k^2 - 28} }{2} = 6$

$\Rightarrow x_2 - x_1 = \dfrac{2 \sqrt{k^2 - 28}}{2} = 6$

$\Rightarrow \sqrt{k^2 - 28} = 6$

Squaring both sides, we get,

$\Rightarrow k^2 - 28 = 36$

$\Rightarrow k = \pm 8$

Hence, the possible values of k are ± 8.