Honors Algebra test review

Simplify

100

3√60

6√15

200

√68x⁷

2x³ √17x

300

6√24 - 2√12

12√6 - 2√3

400

(7-√2)²

51-14√2

500

4x√108x¹¹y³

24x⁶y√3xy

Add/subtract

100

3√64 - 2√ 25

14

200

-3√24 - √6

-7√6

300

√125 + 3√20

11√5

400

√12 - √50 - √27 - √72

-√3 -11√2

500

2√8 -5√32 +2√50

-6√2

Multiply/divide

100

6√2 x 2√6

24√3

200

√6/√24

1/2

300

√8/√5 

(2√10)/5

400

(√72x³y)/(√2x)

6x√y

500

5x³√15x^{4 .} (x²√6x³)

15x⁸√10x

Solve

100

√2x+3 =7

x=23

200

√6x -5=-3

x=2/3

300

√3x-8 =√x+2

x=5

400

3+2√x-1 =15

x=37

500

(√2x²+6)+5=1

no solution

Word problems

100

The legs of a right triangle are 4in and 8in. What is the leg of the hypotenuse in simplest radical form. 

4√5in

200

The hypotenuse of a right triangle is 16m and one of its legs is 14m. What is the length of the other leg in simplest radical form.

2√15m

300

The legs of a right triangle are 2√3cm and 2√6cm. What is the length of the hypotenuse in simplest radical form if necessary.

 6cm

400

FIND THE ERROR!!!

The MISSING LEG of a right triangle with a hypotenuse of √65in and a leg of 7in is 16in. Explain the error and give the correct answer.

To get the correct answer, take the square root of 16 to get an answer of 4in !!!

500

What is the perimeter of the right triangle with the hypotenuse 8√2ft and a leg 4√6ft. Give answer in simplest radical form if necessary.

4√6 + 12√2 ft