Asked by dd
Solve for x in the equation 2 x squared minus 5 x + 1 = 3.
x = five-halves plus-or-minus StartFraction StartRoot 29 EndRoot Over 2 EndFraction
x = five-halves plus-or-minus StartFraction StartRoot 41 EndRoot Over 4 EndFraction
x = five-fourths plus-or-minus StartFraction StartRoot 29 EndRoot Over 2 EndFraction
x = five-fourths plus-or-minus StartFraction StartRoot 41 EndRoot Over 4 EndFraction
All Answers 1
Answered by
GPT-5 mini
AI
First move 3 to the left: 2x^2 − 5x + 1 − 3 = 0 → 2x^2 − 5x − 2 = 0.
Use the quadratic formula x = [5 ± sqrt(b^2 − 4ac)]/(2a) with a=2, b=−5, c=−2:
Discriminant = (−5)^2 − 4(2)(−2) = 25 + 16 = 41.
Thus x = [5 ± sqrt(41)]/(4) = 5/4 ± (sqrt(41))/4.
So the correct choice is: x = five-fourths ± (sqrt(41))/4.
Use the quadratic formula x = [5 ± sqrt(b^2 − 4ac)]/(2a) with a=2, b=−5, c=−2:
Discriminant = (−5)^2 − 4(2)(−2) = 25 + 16 = 41.
Thus x = [5 ± sqrt(41)]/(4) = 5/4 ± (sqrt(41))/4.
So the correct choice is: x = five-fourths ± (sqrt(41))/4.
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