Question

Using the quadratic formula to solve x^2=5-x what are the values of x?

a. -1 +/- √21/2
b.-1 +/-√19i/2
c.5+/- √21/2
d. 1 +/- √19i/2

Answer 1

a
you have to solve x^2+x-5=0
So x=(-1 +/- sqrt(1^2+4*1*5))/2
x=(-1 +/- sqrt(21))/2

Answer 2

Option a -
`x=(-1\pm√(21))/(2)`

Explanation:

Given : Expression
`x^2=5-x`

To find : What are the values of x?

Solution :

The quadratic formula of the quadratic equation
`ax^2+bx+c=0` is

`x=(-b\pm√(b^2-4ac))/(2a)`

Comparing with general equation,
`x^2+x-5=0`

where, a=1 ,b=1, c=-5

Substitute in the formula,

`x=(-b\pm√(b^2-4ac))/(2a)`

`x=(-1\pm√(1^2-4(1)(-5)))/(2(1))`

`x=(-1\pm√(1+20))/(2)`

`x=(-1\pm√(21))/(2)`

Therefore, The value of x is
`x=(-1\pm√(21))/(2)`

So, Option a is correct.