How do you solve #(5x-1)^2=16#?

2 Answers
Aug 25, 2016

Answer:

#x=-3/5# and #x=1#

Explanation:

First you expand #(5x-1)^2# so #(5x-1)*(5x-1) = 25x^2-10x +1#
That leaves us with:
#25x^2-10x +1=16#
Now we subtract 16 from both sides:
#25x^2-10x -15=0#
Then all we have to do is plug a,b and c from #ax^2 +bx+c# into the quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)#
#a=25#
#b=-10#
#c=-15#

#x=(-(-10)+-sqrt((-10)^2-4*25*(-15)))/(2*25)#

#x=(10+-sqrt(100-(-1500)))/50#

#x=(10+-sqrt(1600))/50#

#x=(10+-40)/50#

So #x=50/50=1# and #x=-30/50=-3/5#

Answer:

Take the square root of both sides and solve to get to #x=1#

Explanation:

Start with:

#(5x-1)^2=16#

The first problem we have is that everything on the left is squared. We could multiply it out and end up with #25x^2-10x+1#, but since the right side is a perfect square, let's instead take the square root of both sides:

#sqrt((5x-1)^2)=sqrt16#

#5x-1=4#

And now let's solve:

#5x=5#

#x=1#

And we can check the answer:

#(5x-1)^2=16#

#(5(1)-1)^2=16#

#(5-1)^2=16#

#(4)^2=16#

#16=16#