How do you solve 2x^2 - x - 5 = 02x2−x−5=0 by completing the square?

1 Answer
Leland Adriano Alejandro
Mar 25, 2016

x_1=(1+sqrt(41))/4x1=1+√414
x_2=(1-sqrt(41))/4x2=1−√414

Explanation:

From the given 2x^2-x-5=02x2−x−5=0

Factor out the 2 from the first two terms so that the coefficient of x^2x2 is 1

2(x^2-1/2x)-5=02(x2−12x)−5=0

Now take note of the 1/2 coefficient of x. Divide this 1/2 by 2 it becomes 1/4. Square it, it will become 1/16. This 1/16 will be added and subtracted to the terms inside the grouping symbol.

Let us continue

2(x^2-1/2x)-5=02(x2−12x)−5=0

2(x^2-1/2x+1/16-1/16)-5=02(x2−12x+116−116)−5=0

Notice x^2-1/2x+1/16x2−12x+116 is a PST-Perfect Square Trinomial and
x^2-1/2x+1/16=(x-1/4)^2x2−12x+116=(x−14)2

so that

2((x^2-1/2x+1/16)-1/16)-5=02((x2−12x+116)−116)−5=0

2((x-1/4)^2-1/16)-5=02((x−14)2−116)−5=0

Now, transpose the 5 to the right side then divide both sides by 2 then transpose the 1/16

2((x-1/4)^2-1/16)-5=02((x−14)2−116)−5=0

2((x-1/4)^2-1/16)=52((x−14)2−116)=5

(cancel2((x-1/4)^2-1/16))/cancel2=5/2

(x-1/4)^2-1/16=5/2

(x-1/4)^2=5/2+1/16

Simplify

(x-1/4)^2=41/16

Extract the square root of both sides

sqrt((x-1/4)^2)=+-sqrt(41/16)

x-1/4=+-1/4sqrt(41)

x=1/4+-1/4sqrt(41)

x=(1+-sqrt(41))/4

x_1=(1+sqrt(41))/4
x_2=(1-sqrt(41))/4

God bless...I hope the explanation is useful.

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