Question

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

Answer 1

Answer:

The solutions are:

x1= 5 +√31

x2= 5 -√31

Step-by-step explanation:

We have 6=x^2-10x

Balance the equation by adding the same constant to each side

x^2-10x+25=6+25

x^2-10x+25=31

Rewrite as perfect square,

(x-25)^2=31

Taking square root at both sides

√(x-5)^2 = √31

x-5 = (+/-)√31

x1= 5 +√31

x2= 5 -√31

Therefore the solutions are x1= 5 +√31 , x2= 5 -√31

Answer 2

Answer:

$x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

Step-by-step explanation:

We need to solve the quadratic equation

6 = x^2 -10x

Rearranging we get,

x^2-10x-6=0

Using quadratic formula to solve the quadratic equation

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

a= 1, b =-10 and c=6

Putting values in the quadratic formula

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}$

So, $x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$