Question

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

Answer 1

The solutions are:

x1= 5 +√31

x2= 5 -√31

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

`x = 5+√(31)\,\, and\,\, x=5-√(31)`

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=(-b\pm√(b^2-4ac))/(2a)`

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

Putting values in the quadratic formula

`x=(-(-10)\pm√((-10)^2-4(1)(-6)))/(2(1))\\x=(10\pm√(100+24))/(2)\\x=(10\pm√(124))/(2)\\x=(10\pm√(2*2*31))/(2)\\x=(10\pm√(2^2*31))/(2)\\x=(10\pm2√(31))/(2)\\x = 5\pm√(31)`

So,
`x=5+√(31)\,\, and\,\, x=5-√(31)`