How do you solve using the square root property #4x^2 – 8x – 2 = 0#?

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Answer 1 · 2015-09-24T18:42:28

Refer to explanation

The Square Root Property

For any positive number k, if #x^2 = k# then #x=sqrtk# or #x=-sqrtk#

We want to use the above property to solve the equation given hence we have that

#4x^2-8x-2=0=>4*(x^2-2x+1)-6=0=>(x-1)^2=6/4=>(x-1)^2=3/2#

Hence #(x-1)^2=3/2# using the square root property we get

#(x-1)=sqrt(3/2)=> x=1+sqrt(3/2)#

or

#(x-1)=-sqrt(3/2)=>x=1-sqrt(3/2)#