Question

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

Answer 1

`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`

Answer 2

`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)`