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secondary 3 | A Maths
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Please help me with this question. Thank you
Date Posted: 2 years ago
Views: 392
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done
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clear
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Using quadratic formula
Date Posted:
2 years ago
Thank you very much
squaring both sides is correct method to remove the square root, but squaring both sides may result in "extraneous" solutions which do not satisfy the original equation. so after finding the possible values of x, it is necessary to substitute each value back into original equation to check that the original equation is satisfied.
for this question, on the left-hand side, sqrt[2x] refers only to the positive root. when x = (6+sqrt[11])/25 is substituted into the equation, LHS = +0.8633..., but RHS = -0.8633, so this value of x has to be rejected.
x = (6-sqrt[11])/25 is the only valid answer.
for this question, on the left-hand side, sqrt[2x] refers only to the positive root. when x = (6+sqrt[11])/25 is substituted into the equation, LHS = +0.8633..., but RHS = -0.8633, so this value of x has to be rejected.
x = (6-sqrt[11])/25 is the only valid answer.
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