Question

Using the quadradic formula to solve x^2=5-x, what are the values of x

Answer 1

x1 = 2.79129, x2 = 1.79129

Explanation:

x^2-5-x

x^2-5+x=0

x^2+x-5=0

-1+-square root od=f 1^2 - 4x1x(-5) / 2x1 = -1+- square root of 21 (add numbers together) / 2

then solve the formula with a plus sign instead of +- then and solve the formula with a - this time and you should get x1 = 2.79129, x2 = 1.79129, or -1 +square root of 21 (add numbers together) / 2 and -1 - square root of 21 (add numbers together) / 2

Answer 2

For this case we must solve the following equation:

`x ^ 2 = 5-x`

By manipulating algebraically we have:

`x ^ 2 + x-5 = 0`

The quadratic formula is given by:

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

We have to:

`a = 1\\b = 1\\c = -5`

Substituting we have:

`x = \frac {-1 \pm \sqrt {1 ^ 2-4 (1) (- 5)}} {2 (1)}\\x = \frac {-1 \pm \sqrt {1 + 20}} {2}\\x = \frac {-1 \pm \sqrt {21}} {2}`

So, we have two roots:

`x_ {1} = \frac {-1+ \sqrt {21}} {2} = 1,7913\\x_ {2} = \frac {-1- \sqrt {21}} {2} = - 2.7913`

Answer:

`x_ {1} = \frac {-1+ \sqrt {21}} {2}\\x_ {2} = \frac {-1- \sqrt {21}} {2}`