Algebra Honors ( Periods 5 & 6)

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


Fractional Exponents

In chapter 4 we reviewed the law of exponents:
a^m ⋅aⁿ = a^m+n
Thus you know
2⁴⋅2⁵= 2⁹
What do you notice? What would be the value of n in the equation
2ⁿ⋅2ⁿ = 2
Using what we know from above,
2ⁿ⋅2ⁿ = 2ⁿ⁺ⁿ = 2²ⁿ

The bases are equal ( and NOT -1, 0 or 1). Therefore the exponents must be equal.
That says
2n = 1
n = 1/2

and you have
2^1/2⋅2^1/2=2
Because √2⋅√2 = 2 and (-√2)(-√2) = 2 we note that 2^1/2 as either the positive or negative square root of 2
Selecting the positive or principal square root we define,
2^1/2 = √2

Radicals are not restricted to square roots. The symbol ∛ represents the third ( or cube) root, ∜ represents the fourth root and so on...
As you have learned the root index is omitted when n = 2

Just as the inverse of squaring a number is finding the square root, the inverse of cubing a number is finding the cube root. Since 2³ = 8
∛8 ( read the cube root of 8) is 2.
Likewise (-2)³ = -8
∛(-8) = -2

BE CAREFUL---> While ∛-8 is a real number √-8 is not
In general, you CAN find ODD roots of negative numbers but not EVEN Roots!!

Solve
4ⁿ⋅4ⁿ⋅4ⁿ= 4
4³ⁿ = 4
Since the bases are EQUAL ( that's the KEY), the exponents are also!!
so 3n = 4
n = 3/4


You know that ∛7 = 7 ^1/3 So How would you write (∛7) ² in exponential form?
(∛7) ² = (7^1/3)² = 7^(1/3)2 = 7^2/3

Simplify:

16^3/4
First write as
∜16³
Now change 16 into 2⁴ Why?
You end up with ∜(2⁴)³
Looking at just ∜2⁴ you realize you have 2
and so you are left with
2³ = 8