Radical Equations - A radical equation is an equation which contains a variable in a radical. We recall that the inverse (or opposite) operation of "Square Root" is "Squaring".
Ex. (√3)^2 = 3, (√x+4)^2 = x+4
Remember this: To solve a radical equation, isolate the radical and then square it.
Note: When solving radical equations, you must check the answers that you get since during the process of solving the equations you may have created an extraneous root.
Equations with One Radical term!
5 + √2x-1 = 12
Step 1! State the restrictions on the variable "x" LIKE A BOSS.
2x - 1 = ≥ 0
2x = ≥ 1
x = ≥ 1/2
Step 2! Solve that equation LIKE A BOSS.
5 + √2x-1 = 12
(bring 5 to the other side of the equal sign,)
√2x-1 = 12-5
√2x-1 = 7
(now square both sides!)
(√2x-1)^2 = (7)^2
2x - 1 = 49
2x/2 = 50/2 (Remember that the 2 cancels out so you're only left with "x").
x = 25!
So you solved the equation LIKE A BOSS, but the final step is to check your equation!
5 + √2x-1 = 12
*Substitute your 25 in for x*
5 + √2(25)-1 = 12
5 + √49 = 12
5 + 7 = 12
12 = 12 L=R
You just solved a Radical equation with one radical term LIKE A BOSS!
Now that you have solved a Radical equation with one radical term.. now you will learn how to solve Radical Equations w/ an Extraneous Root LIKE A KING OR A QUEEN!
Note: To help solve Radical Equations w/ an Extraneous Root LIKE A KING OR A QUEEN, remember this equation!
(a+b)^2 = a^2 + 2ab + b^2
ex. n-√5-n = -7
5 - n ≥ 0
5 ≥ n
n - √5-n = -7
(take the -7 to one side and take the √5-n to the other side)
(n+7)^2 = (√5-n)^2
Note: this is where you use the equation a^2 + 2ab +b^2 comes in! Square your right side, then use the equation up here ^^^^^^^ to solve for the left side.
n^2+14n+49 = 5-n
Take your (5-n) and bring it to the other side!
n^2+14n+n+49-5=0
Combine like terms LIKE A KING OR QUEEN.
n^2 +15n+44=0
(n+11)(n+4) =0
n +11= 0 , n +4 = 0
n= -11 , n= -4
we have solved the equation LIKE A KING OR QUEEN! Time to check our answers.
n -√5-n = 7
-11 -√5-(-11) = 7
-11 -√16 = 7
-11 - 4 = 7
-15 = 7 it doesn't work!
let's try -4!
n-√5-n = 7
-4 -√5-(-4) = 7
-4 -√9 = 7
-4 - 3 = 7
7 = 7
Therefore, only -4 works!
The last equation you have to solve is to solve an equation with two radicals.. LIKE A GOD.
When you are solving, isolate 1 radical and then square both sides.
ex. 7 + √3x = √5x+4 +5 , x ≥ 0
5x+4 ≥ 0
5x ≥ -4
x ≥ -4/5
7-5 + √3x = √5x+4
(2 + √3x)^2 = (√5x+4)^2
4+4√3x + 3x = 5x + 4
4 √3x = 5x +4 3x-4
(4√3x)^2 = (2x)^2
16 ● 3x = 4x^2
4x^2 - 48x = 0
4x (x-12) = 0
4x =0 or x-12=0
x= 0 or x= 12
So much work, but we solved the equation.. LIKE A GOD I THINK SO! Time to check! :D
7 +√3x = √5x+4 +5
7 + √3(0) = √5(0)+4 +5
7 + 0 = 2 + 5
7 = 7
We shall now try using 12!
7 + √3(12) = √5(12)+4 +5
7 + 6 = 8 + 5
13 = 13
YAY! We solved Radical Equations LIKE A BOSS, we solved Radical Equations w/ an Extraneous Root LIKE A KING AND QUEEN, and lastly, we solved Radical Equations with two radicals LIKE A GOD.
I hope all this helped! Sorry if it was too long (twss.. hehe.) Anyways, see you all Monday! :D
-Neil Cymbalisty.. OUT!
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