How do you solve #32x^2-3x-14=(2x-1)^2#?

2 Answers
May 20, 2018

Answer:

#x=5/7 orx = -3/4#

Explanation:

#32x^2 -3x-14 = (2x-1)^2" "larr# remove the brackets

#32x^2 -3x-14 = 4x^2 -4x+1" "larr# make it #=0#

#32x^2-4x^2-3x+4x-14-1=0#

#28x^2+x-15 =0" "larr# factorise the quadratic

#(7x-5)(4x+3)=0" "larr# solve each factor set #=0#

If #7x-5 = 0 rArr" "x = 5/7#

If #4x+3=0 rArr" "x = -3/4#

May 20, 2018

Answer:

#x = 5/7, x = -3/4#

Explanation:

#32x^2-3x-14=(2x-1)^2# expand the right side,
#32x^2-3x-14=4x^2-4x+1#
#32x^2-4x^2-3x+4x-14-1=0# shift them to the left side
#28x^2+x-15=0#

by using the equation #(-b+- sqrt(b^2-4ac))/(2a)#

#a=28, b=1, c=-15#

sub a,b,c into the equation#(-b+- sqrt(b^2-4ac))/(2a)#

#x=(-1+- sqrt(1^2-4(28)(-15)))/(2(28))#
#x=(-1+- sqrt(1+1680))/(56)#
#x=(-1+- sqrt(1681))/(56)#
#x=(-1+- 41)/(56)#
#x=(-1+ 41)/(56)# or #x=(-1- 41)/(56)#
#x=( 40)/(56)# or #x=(-42)/(56)#

#x=5/(7)# or #x=(-3)/(4)#