How do you solve y=5x2+20x+23 using the completing square method?

1 Answer
George C.
Dec 27, 2016

x=−2±√y−35

Explanation:

I will assume that we want to solve the equation in the sense of expressing possible values of x in terms of y.

If you actually just wanted to know the zeros, we can then put y=0 and evaluate the result.

Given:

y=5x2+20x+23

We will perform a sequence of steps to isolate x on one side.

First divide both sides by 5 to make the coefficient of x2 into 1:

y5=x2+4x+235

If we take half of the coefficient of x then we get the value 2, so note that:

(x+2)2=x2+4x+4

which matches the right hand side of our equation in the first two terms.

So we can proceed as follows:

y5=x2+4x+235

y5=x2+4x+4−4+235

y5=(x+2)2+35

Subtracting 35 from both ends we get:

y−35=(x+2)2

Transposed:

(x+2)2=y−35

Take the square root of both sides, allowing for the possibility of either sign of square root to get:

x+2=±√y−35

Finally subtract 2 from both sides to find:

x=−2±√y−35

If we put y=0 to find the zero of the quadratic then we get a pair of non-Real Complex values:

x=−2±√0−35=−2±√−35=−2±√35i=−2±√155i

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