How do you solve 5x^2+3x-4=05x2+3x−4=0 by completing the square?

1 Answer
Leland Adriano Alejandro
Mar 27, 2016

The roots are

x_1=(-3+sqrt(89))/10x1=−3+√8910
x_2=(-3-sqrt(89))/10x2=−3−√8910

Explanation:

From the given
5x^2+3x-4=05x2+3x−4=0

factor out 5 from the first two terms

5(x^2+3/5x)-4=05(x2+35x)−4=0

Use now the number 3/5. Divide this number by 2 then square the result to obtain 9/100. This 9/100 will be added and subtracted to terms inside the grouping symbol.

Let us continue

5(x^2+3/5x)-4=05(x2+35x)−4=0

5(x^2+3/5x+9/100-9/100)-4=05(x2+35x+9100−9100)−4=0

We now have a perfect square trinomial

x^2+3/5x+9/100=(x+3/10)^2x2+35x+9100=(x+310)2 so that

5((x+3/10)^2-9/100)-4=05((x+310)2−9100)−4=0

Transpose now the -4 to the right side then divide by 5 both sides of the equation then transpose the -9/100 to the right also.

5((x+3/10)^2-9/100)=45((x+310)2−9100)=4

(cancel5((x+3/10)^2-9/100))/cancel5=4/5

(x+3/10)^2-9/100=4/5

(x+3/10)^2=4/5+9/100

simplify

(x+3/10)^2=(80+9)/100

(x+3/10)^2=89/100
Extract the square root of both sides

sqrt((x+3/10)^2)=+-sqrt(89/100)

x+3/10=+-sqrt(89/100)

x=-3/10+-sqrt(89)/10

The roots are

x_1=-3/10+sqrt(89)/10
x_2=-3/10-sqrt(89)/10

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

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