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Powers and Roots Practice Test 3

Question : 16 of 40

Marks: +1, -0

Find x, if

\sqrt{1+\frac{97}{2304}}=1+\frac{x}{48}.

Solution:

Concept:

Simplify the square root and compare the numerators after making denominators equal.

Explanation:

Given:

\sqrt{1+\frac{97}{2304}}=1+\frac{x}{48}

Simplify inside the square root:

1+\frac{97}{2304}=\frac{2304+97}{2304}=\frac{2401}{2304}

Now take the square root:

\sqrt{\frac{2401}{2304}}=\frac{49}{48}

Rewrite the right side with denominator

48

1+\frac{x}{48}=\frac{48+x}{48}

Equate both sides:

\frac{49}{48}=\frac{48+x}{48}

Cancel the common denominator:

49=48+x

Solve for

x

x=49-48=1

Answer:

x=1

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