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The solution of the equation \( \Large \sqrt{25 - x^{2}} = x - 1 \) are
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A) x = 3 and x = 4 |
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B) x = 5 and x = 1 |
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C) x = -3 and x = 4 |
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D) x = 4 and x \( \Large \neq \) -3 |
Correct Answer:
| D) x = 4 and x \( \Large \neq \) -3 |
Description for Correct answer:
\( \Large \sqrt{25 - x^{2}} = x - 1 \)
=> \( \Large 25 - x^{2} = (x - 1)^{2} \)
=> \( \Large 25 - x^{2} = x^{2} + 1 - 2x \)
=> \( \Large 2x^{2} - 2x - 24 = 0 \)
=> \( \Large x^{2} - x - 12 = 0 \)
=> (x - 4)(x + 3) = 0
=> x = 4, x = -3
\( \Large \therefore x = 4 \)
\( \Large ( \because x = -3) \) does not satisfy the given equation.
\( \Large \sqrt{25 - x^{2}} = x - 1 \)
=> \( \Large 25 - x^{2} = (x - 1)^{2} \)
=> \( \Large 25 - x^{2} = x^{2} + 1 - 2x \)
=> \( \Large 2x^{2} - 2x - 24 = 0 \)
=> \( \Large x^{2} - x - 12 = 0 \)
=> (x - 4)(x + 3) = 0
=> x = 4, x = -3
\( \Large \therefore x = 4 \)
\( \Large ( \because x = -3) \) does not satisfy the given equation.
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