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Powers and Roots Practice Test 4

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Question : 4 of 40

Marks: +1, -0

Find x in:

\sqrt{3364}+\sqrt{40+x^3}=74 .

5
-4
-6
6
-5

Solution:

Concept:

A radical equation is solved by isolating the square root and then squaring both sides.

Explanation:

Given equation:

\sqrt{3364}+\sqrt{40+x^3}=74

.

First, find

\sqrt{3364}

.

Since

58^2=3364

, we get

\sqrt{3364}=58

.

Substitute this value.

58+\sqrt{40+x^3}=74

.

Isolate the radical term.

\sqrt{40+x^3}=74-58=16

.

Square both sides.

40+x^3=16^2=256

.

Subtract

40

from both sides.

x^3=216

.

Take the cube root.

x=\sqrt[3]{216}=6

.

Check:

58+\sqrt{40+216}=58+\sqrt{256}=58+16=74

.

So the solution is correct.

Answer:

x=6

, which is option D.

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