How do you solve #x^ { 2} + 2x - 28= 0#?

2 Answers
Shwetank Mauria
Dec 25, 2016

#x=-1+sqrt29# or #-1-sqrt29#

Explanation:

In the given equation#x^2+2x-28=0#, the discriminant #b^2-4ac=2^2-4xx1xx(-28)=4+112=116# is not a perfect square. There this cannot be solved by splitting middle term.

Hence we use quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)#

Hence #x=(-2+-sqrt116)/2#

= #(-2+-2sqrt29)/2#

= #-1+sqrt29# or #-1-sqrt29#

George C.
Dec 25, 2016

#x = -1+-sqrt(29)#

Explanation:

This difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We will use this later with #a=(x+1)# and #b=sqrt(29)#

Given:

#x^2+2x-28 = 0#

Notice that #x^2+2x# is the first part of the square trinomial:

#x^2+2x+1 = (x+1)^2#

So we can complete the square like this:

#0 = x^2+2x-28#

#color(white)(0) = x^2+2x+1-29#

#color(white)(0) = (x+1)^2-(sqrt(29))^2#

#color(white)(0) = ((x+1)-sqrt(29))((x+1)+sqrt(29))#

#color(white)(0) = (x+1-sqrt(29))(x+1+sqrt(29))#

Hence:

#x = -1+-sqrt(29)#

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