Radical Functions & Equations — Challenge
Squaring both sides is legal, and it can still hand you an answer that is wrong.
What these problems cover
These questions use 5 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Isolate — Get the variable alone on one side.
- Square root — For k ≥ 0, √k is the nonnegative number whose square is k.
- Radical — The √x symbol, and the expression under it.
- Roots — The x values where y = 0.
- Rational exponent — x1/2 = √x and xm/n = √[n]xm.
Problem 1
Solve √x=5.
- 5
- 10
- 25
- 2.5
- 625
Worked answer
25
Square both sides to undo the square root: x = 52.
x = 25.
Check: √25 = 5, so the solution is genuine.
Problem 2
Solve √x+1=3.
- 2
- 8
- 9
- 10
- 26
Worked answer
8
Square both sides: x+1 = 9.
Subtract 1: x = 8.
Check: √8+1 = √9 = 3.
Problem 3
Solve √x+2=7.
- 5
- 25
- 45
- 49
- 9
Worked answer
25
Isolate the root first: subtract 2 to get √x = 5.
Then square both sides: x = 25.
Squaring before isolating would give x+4 = 49, which is wrong because (√x+2)2 is not x+4.
Problem 4
Solve √2x=6.
- 3
- 12
- 18
- 36
- 72
Worked answer
18
Square both sides: 2x = 36.
Divide by 2: x = 18.
Check: √2· 18 = √36 = 6.
Problem 5
How many real solutions does √x+3=x-3 have?
- 0
- 1
- 2
- 3
- Infinitely many
Worked answer
1
Square both sides: x+3=x2-6x+9, so x2-7x+6=0 and (x-1)(x-6)=0. Check x=1: √4=2 but 1-3=-2. It fails. Check x=6: √9=3 and 6-3=3. It works. Only one solution survives.
Problem 6
What is the value of √6+√6+√6+⋯, continuing forever?
- 2
- 3
- 4
- 6
- √6
Worked answer
3
Call the whole expression y. The part under the first root is 6 plus the same infinite expression, so y=√6+y. Square: y2=y+6, so y2-y-6=0 and (y-3)(y+2)=0. A square root is not negative, so y=3.
Problem 7
If x3/2=27, what is x?
- 3
- 9
- 27
- 81
- 243
Worked answer
9
Raise both sides to the power 23: x=272/3. Cube root first: √[3]27=3. Then square: 32=9. Check: 93/2=(√9)3=27.
Problem 8
What is the sum of all solutions of √2x+7-√x+3=1?
- -2
- 1
- 2
- -3
- 4
Worked answer
-2
Move one root across: √2x+7=1+√x+3, then square. 2x+7=1+2√x+3+x+3, so x+3=2√x+3. With u=√x+3 this is u2=2u, giving u=0 or u=2, so x=-3 or x=1. Both check out, and -3+1=-2.