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belka
14 days ago
17

What are the solutions to the quadratic equation (5y + 6)2 = 24? y = StartFraction negative 6 + 2 StartRoot 6 EndRoot Over 5 End

Fraction and y = StartFraction negative 6 minus 2 StartRoot 6 EndRoot Over 5 EndFraction y = StartFraction negative 6 + 2 StartRoot 6 EndRoot Over 5 EndFraction and y = StartFraction 6 minus 2 StartRoot 6 EndRoot Over 5 EndFraction y = StartFraction negative 4 StartRoot 6 EndRoot Over 5 EndFraction and y = StartFraction negative 8 StartRoot 6 EndRoot Over 5 EndFraction y = StartFraction 4 StartRoot 6 EndRoot Over 5 EndFraction and y = StartFraction 8 StartRoot 6 EndRoot Over 5 EndFraction
Mathematics
2 answers:
Leona [12.6K]14 days ago
10 0

Response:

The equation yields two distinct solutions:

a) y=\frac{-6+2\sqrt{6} }{5}, and

b) y=\frac{-6-2\sqrt{6} }{5}

Detailed breakdown:

Given the equation: (5y+6)^2=24, since the perfect square involving the variable "y" is positioned on the left side of the equality, we will proceed by taking the square root of both sides, followed by isolating the variable:

(5y+6)^2=24\\\sqrt{(5y+6)^2} =+/-\sqrt{24} \\(5y+6)=+/-\sqrt{6*4} \\(5y+6)=+/-2\sqrt{6}\\5y=-6+/-2\sqrt{6}\\y=\frac{-6+/-2\sqrt{6} }{5}

Consequently, there are two solutions:

a) y=\frac{-6+2\sqrt{6} }{5}, and

b) y=\frac{-6-2\sqrt{6} }{5}

Svet_ta [12.7K]14 days ago
5 0

Response: A

y=\frac{-6+2\sqrt{6} }{5} \\y=\frac{-6-2\sqrt{6} }{5}

Detailed explanation:

just performed this

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