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Teachers & SchoolsTo solve radical equations, you square both sides, FOIL it out if necessary, and simplify. Ta da!

Algebra | Exponents and Radicals |

Language | English Language |

Mathematics and Statistics Assessment | Powers, Roots, and Radicals |

plus 6 solve for X. First sqaure both sides to get rid of the square roots

leaving us with 3 x minus 2 equals x plus 6. Next add 2 to both sides and subtract [Chainsaw chopping through a log]

X from both sides, we get 2x equals 8 divide both sides by 2 and boom x equals 4.

The next one's a little more complicated, square root of 3x plus 4 equals x minus 5. [Next problem written on a post-it note]

Square both sides to get rid of the square root on the left side and we get

3x plus 4 equals X minus 5 in parentheses squared. Now we're gonna

use foil on the right side, we can start by multiplying X by X which gives us x [Tin foil wrapped into a ball]

squared then we multiply X by minus 5 and we get -5x. We continue by

multiplying minus 5 by X we get minus 5x and we finish by multiplying minus 5 by

minus 5 which equals positive 25, our new equation now reads 3x plus 4 equals x

squared minus 10x plus 25. Subtract 3x and 4 from both sides and we

get x squared minus 13x plus 21 equals 0. Using the quadratic formula to solve for

X is shown as x equals negative b plus or minus the square root of b squared [Man looks excited and says 'Look a cheatsheet!']

minus 4ac (breathes in) all over 2a. We see that X equal 13 plus or minus the square root

of negative 13 squared minus 4 times 21 times 1 over 2. This simplifies into x

equals 13 plus or minus the square root of 169 minus 84 over 2. Therefore the

answer is x equals 13 plus or minus the square root of 85 over 2. Now it can't be

reduced any further just like your friends interest in going to this gig. [Woman looks happy that the problem is solved]