How do you solve using the completing the square method #x^2 - 5x + 1 = 0#?

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Answer 1 · 2016-05-30T12:13:19

#x^2-5x + 1 =(x-5/2-sqrt(21)/2)(x-5/2+sqrt(21)/2)#

The completing square method is a way of solving quadratic polynomials by reducing then to the structure

#(x+alpha)^2-beta^2 equiv (x+alpha + beta)(x+alpha-beta)#

The right one is the factorized form.

Now giving a polynomial #p_2(x)=x^2+a_1x+a_0# the reduction begins by making

#x^2+a_1x+a_0 = (x+alpha)^2-beta^2#.

That furnishes

#{(a_1 = 2 alpha),(a_0=alpha^2-beta^2):}#

Taking #alpha = a_1/2# we obtain #beta = pm sqrt((a_1/2)^2-a_0)# so the result is

#x^2+a_1x+a_0 equiv (x+a_1/2)^2-(sqrt((a_1/2)^2)-a_0)^2#

If #(a_1/2)^2 -a_0< 0# the roots are complex conjugate.
In the present case we have

#x^2-5x + 1 = (x-5/2)^2 - (sqrt((5/2)^2)-1)^2#

#x^2-5x + 1 =(x-5/2-sqrt(21)/2)(x-5/2+sqrt(21)/2)#