Using the quadratic formula to solve x2 = 5 – x, what are the values of x?

Asked by maham237 @ in Mathematics viewed by 367 People

Answered by maham237 @


Answer:

A quadratic equation ax^2+bx+c = 0 ....[1] where, a, b and c are coefficient.

Then the solution is given by:

x = -\frac{b \pm \sqrt{b^2-4ac}}{2a} ....[2]

As per the statement:

Given the equation:

x^2=5-x

⇒x^2+x-5 = 0

On comparing with [1] we have;

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

Then substitute these in [2] we have;

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

Therefore, the values of x are:

x = \frac{-1 + \sqrt{21}}{2} , \frac{-1 - \sqrt{21}}{2}


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