Algebra Honors (Period 6 & 7)

Simple Radical Equations 11-10

Solving equations involving radicals are solved by isolating the radical on one side of the equals sign and then squaring both sides of the equation.
140 = √2(9.8)d all under the √
140 = √19.6d
(140)² = (√19.6d)²
19600 = 19.6d
1000=d
The solution set is {1000}
Solve
√(5x+1) + 2 = 6
√(5x+1) = 4
[√(5x+1)]² = (4)²
5x + 1 = 16
5x = 15
x = 3
The solution set is {3}

When you square both sides of an equation, the new equation may NOT be equivalent to the original equation Therefore, you must CHECK EVERY POSSIBLE ROOT IN THE ORIGINAL EQUATION to see whether it is indeeed a root.
Solve
√(11x² -63) - 2x = 0

√(11x² -63) = 2x

√(11x² -63)² = (2x)²
11x² -63 = 4x²
7x² = 63
x² = 9
x = ± 3
Now we need to check for BOTH + 3 and - 3

Rewrite the original equation



√(11x² -63) - 2x = 0
√(11(3)² -63) - 2(3) = 0
√99-63 - 6 = 0
√36 - 6 = 0
6-6 = 0
That's true
Now for x = -3

√(11(-3)² -63) - 2(-3) = 0
√(99 -63) + 6 = 0
√36 + 6 = 0
12 ≠ 0
So -3 is NOT a solution