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

1 Answer

Answer:

The roots are

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

Explanation:

From the given
#5x^2+3x-4=0#

factor out 5 from the first two terms

#5(x^2+3/5x)-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=0#

#5(x^2+3/5x+9/100-9/100)-4=0#

We now have a perfect square trinomial

#x^2+3/5x+9/100=(x+3/10)^2# so that

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