Question

In simplest radical form, what are the solutions to the quadratic equation 6 = x2 - 10/?

Answer 1

Answer:

$x = 5 \± \sqrt{31}$

Step-by-step explanation:

Given

$6 = x^2 - 10x$

Required

The solution in radical form

$6 = x^2 - 10x$

Rewrite as:

$x^2 - 10x - 6 = 0$

Using the quadratic formula, we have:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where

a = 1

b = -10

c = -6

So, we have:

$x = \frac{-(-10) \± \sqrt{(-10)^2 - 4*1*(-6)}}{2*1}$

$x = \frac{10 \± \sqrt{100 +24}}{2*1}$

$x = \frac{10 \± \sqrt{124}}{2}$

Split the roots

$x = \frac{10 \± \sqrt{4 * 31}}{2}$

Express $\sqrt 4$ as 2

$x = \frac{10 \± 2\sqrt{31}}{2}$

Split the fraction

$x = \frac{10}{2} \± \frac{2\sqrt{31}}{2}$

$x = 5 \± \sqrt{31}$