Asked by kelvin calin
Which is the solution of the quadratic equation (4y – 3)2 = 72?
y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction and y = StartFraction 3 minus 6 StartRoot 2 EndRoot Over 4 EndFraction
y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction and y = StartFraction negative 3 minus 6 StartRoot 2 EndRoot Over 4 EndFraction
y = StartFraction 9 StartRoot 2 EndRoot Over 4 EndFraction and y = StartFraction negative 3 StartRoot 2 EndRoot Over 4 EndFraction
y = StartFraction 9 StartRoot 2 EndRoot Over 4 EndFraction and y = StartFraction 3 StartRoot 2 EndRoot Over 4 EndFraction
All Answers 1
Answered by
GPT-5 mini
AI
Solve (4y − 3)^2 = 72.
Take square roots: 4y − 3 = ±√72 = ±6√2.
Thus 4y = 3 ± 6√2, so y = (3 ± 6√2)/4.
Therefore the correct choice is: y = (3 + 6√2)/4 and y = (3 − 6√2)/4.
Take square roots: 4y − 3 = ±√72 = ±6√2.
Thus 4y = 3 ± 6√2, so y = (3 ± 6√2)/4.
Therefore the correct choice is: y = (3 + 6√2)/4 and y = (3 − 6√2)/4.
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