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Nadusha1986 [10]
3 years ago
8

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

Mathematics
2 answers:
kenny6666 [7]3 years ago
3 0
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
Law Incorporation [45]3 years ago
3 0

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.

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