Solving Radicals

Question

square root of quantity (7t+2) equals 2t

_/(7t+2)=2t

Answer 1

No one can mess with the expression on the left because there's a radical in the way. Get rid of that, by doing the opposite of square rooting, which is squaring.

 

[√(7t + 2)]² = (2t)²

7t + 2 = 4t²

 

What we have here is a quadratic equation. Quadratics need to be set equal to zero to be solved.

 

7t + 2 = 4t²

4t² - 7t - 2 = 0

 

Here's how to factor these. Multiply the a and c terms together. Those terms are 4 and -2, so their product is -8. The b term is -7. To factor this, find two numbers that add up to -7 but multiply to -8. Those numbers are -8 and positive 1. This is what to do with the magic numbers.

 

(4t - 8)(4t + 1) = 0

 

The 4's in front of the t's came from the coefficient of t² in the equation. Simplify the factoring by dividing by a GCF in each parenthetical term. (4t - 8) has a GCF of 4 in it, so divide those terms by 4. In (4t + 1), there's no GCF so leave it alone.

 

(4t - 8)(4t + 1) = 0

(t - 2)(4t + 1) = 0

 

This is now the product of two terms, and they multiply to be zero. That means each term can be zero to make this equation true. Solve for t by setting each factor equal to zero.

 

t - 2 = 0

t = 2

 

4t + 1 = 0

t = -1/4

 

There's one last step. In square root functions, when there's a variable inside the radical and also one outside, there's a chance there might be "extraneous solutions." That means that one of this is a "fake" solution that might not work. To see which one's fake, plug them back in. Start with t = -1/4.

 

√(7t + 2) = 2t

√(7(-1/4) + 2) = 2(-1/4)

√(1/4) = -1/2

1/2 ≠ -1/2

 

This solution does not work. Hopefully t = 2 will work.

 

√(7t + 2) = 2t

√(7(2) + 2) = 2(2)

√16 = 4

4 = 4

 

Looks like t = 2 works after all. This is the one and only solution. I hope this helps, and good luck with the rest of your math class!