Asked by kasdfa
What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve.
x = StartFraction negative 4 plus-or-minus StartRoot 5 EndRoot Over 2 EndFraction
x = StartFraction negative 7 plus-or-minus StartRoot 5 EndRoot Over 2 EndFraction
x = –7 and x = –2
x = –1 and x = 4
All Answers 1
Answered by
GPT-5 mini
AI
Let u = 2x + 3. Then the equation becomes
u^2 + 8u + 11 = 0.
Use the quadratic formula: u = [−8 ± sqrt(8^2 − 4·1·11)]/2 = [−8 ± sqrt(64 − 44)]/2 = [−8 ± sqrt(20)]/2
= −4 ± sqrt(5).
Now substitute back: 2x + 3 = −4 ± sqrt(5) ⇒ 2x = −7 ± sqrt(5) ⇒ x = (−7 ± sqrt(5))/2.
So the correct choice is x = (−7 ± √5)/2.
u^2 + 8u + 11 = 0.
Use the quadratic formula: u = [−8 ± sqrt(8^2 − 4·1·11)]/2 = [−8 ± sqrt(64 − 44)]/2 = [−8 ± sqrt(20)]/2
= −4 ± sqrt(5).
Now substitute back: 2x + 3 = −4 ± sqrt(5) ⇒ 2x = −7 ± sqrt(5) ⇒ x = (−7 ± sqrt(5))/2.
So the correct choice is x = (−7 ± √5)/2.
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