Question

n simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x? Quadratic formula: x = x = 5 x = 5 x = 5 x = 5

Answer 1

For this case we must find the solutions of the following quadratic equation:

`6 = x ^ 2-10x\\x ^ 2-10x-6 = 0`

Where:

`a = 1\\b = -10\\c = -6`

According to the quadratic equation we have:

`x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\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 {124}} {2}\\x = \frac {10 \pm \sqrt {31 * 4}} {2}`

`x = \frac {10 \pm 2\sqrt {31}} {2}`

So, we have two roots:

`x_ {1} = 5 + \sqrt {31}\\x_ {2} = 5- \sqrt {31}`

ANswer:

`x_ {1} = 5 + \sqrt {31}\\x_ {2} = 5- \sqrt {31}`

Answer 2

`x=5+√(31)` or
`x=5-√(31)`

Explanation:

The given quadratic equation is

`6=x^2-10x`

Rewrite in the form;

`ax^2+bx+c=0`

This implies that;

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

This gives us a=1,b=-10,c=-6

We use the quadratic formula;

`x=(-b\pm√(b^2-4ac) )/(2a)`

Plug in the values to get;

`x=(--10\pm √((-10)^2-4(1)(-6)) )/(2(1))`

`x=(10\pm √(100+24) )/(2)`

`x=(10\pm √(124) )/(2)`

`x=(10\pm2√(31) )/(2)`

`x=5\pm√(31)`

`x=5+√(31)` or
`x=5-√(31)`