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UPSC CDS 2 2023 Math Paper

Question : 11 of 100

Marks: +1, -0

\sqrt{x+9}=x-3

have?

Solution:

Concept:

A square root equation must be solved by squaring both sides and checking for extraneous roots.

Explanation:

Start with

\sqrt{x+9} = x - 3

Square both sides:

x + 9 = (x - 3)^2

Expand:

x + 9 = x^2 - 6x + 9

Simplify:

x^2 - 7x = 0

Factor:

x(x - 7) = 0

x = 0

x = 7

Check each root in the original equation.

For

x = 0

: LHS

\sqrt{0+9}=3

, RHS

0-3=-3

→ not equal, discard.

For

x = 7

: LHS

\sqrt{7+9}=4

, RHS

7-3=4

→ works.

Only one real root exists.

Answer:

Only one real root.

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