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

Answer:

Option a - $x=\frac{-1\pm\sqrt{21}}{2}$

Step-by-step 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

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

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

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

Substitute in the formula,

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

$x=\frac{-1\pm\sqrt{1^2-4(1)(-5)}}{2(1)}$

$x=\frac{-1\pm\sqrt{1+20}}{2}$

$x=\frac{-1\pm\sqrt{21}}{2}$

Therefore, The value of x is $x=\frac{-1\pm\sqrt{21}}{2}$

So, Option a is correct.