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

2 Answers
Ricardo A.
Aug 25, 2016

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

MathFact-orials.blogspot.com
Aug 25, 2016

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

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