Question

Solve the quadratic equation (y²-6y)²-4(y²-6y+3)-20=0​

Answer 1

(y^2 -6y)^2 - 4(y^2 -6y + 3) - 20 = 0

Let y^2 -6y = x ——> 1

x^2 - 4(x + 3) - 20 = 0

x^2 - 4x - 12 - 20 = 0

x^2 - 4x - 32 = 0

x^2 - 8x + 4x - 32 = 0

x(x - 8) + 4(x - 8) = 0

(x + 4)(x - 8) = 0

x + 4 = 0, x - 8 = 0

x = -4, x = 8

Substitute x = -4 in equation 1,

y^2 -6y = -4

y^2 -6y + 4 = 0

y = (-(-6) +/- sqrt(36 - 16))/2

y = (6 +/- sqrt(20))/2

y = (6 +/- 2*sqrt(5))/2

y = 3 +/- sqrt(5)

y = 3 + sqrt(5), 3 - sqrt(5)

Substitute x = 8 in equation 1,

y^2 -6y = 8

y^2 -6y - 8 = 0

y = (-(-6) +/- sqrt(64 + 32))/2

y = (-(-6) +/- sqrt(96))/2

y = (-(-6) +/- 4*sqrt(6))/2

y = 3 +/- 2*sqrt(6)

y = 3 + 2*sqrt(6), 3 - 2*sqrt(6)

Therefore

y = 3 + sqrt(5), 3 - sqrt(5), 3 + 2*sqrt(6), 3 - 2*sqrt(6) ——> Answer

Answer 2

On solving, y = 3 ± √5, y = 7, and y = - 1.

Given,

A quadratic equation, (y² - 6y)² - 4 (y² - 6y + 3) - 20 = 0

To Find,

The solution of the given quadratic equation.

Solution,

We have, (y² - 6y)² - 4 (y² - 6y + 3) - 20 = 0

Let (y² - 6y) be x.

⇒ x² - 4(x + 3) - 20 = 0

⇒ x² - 4x -12 -20 = 0

⇒ x² - 4x - 32 = 0

⇒ x² - 8x + 4x - 32 = 0

⇒ x(x - 8) + 4(x - 8) = 0

⇒ (x + 4)(x - 8) = 0

⇒ x = - 4 and x = 8

Now, substituting the values which are let,

⇒ (y² - 6y) = x

For x = - 4,

⇒ y² - 6y + 4 = 0

⇒ y = [6 ± √(36 - 16)]/2

⇒ y = [6 ± 2√5]/2

⇒ y = 3 ± √5

For x = 8,

⇒ y² - 6y - 8 = 0

⇒ y = [6 ± √(36 + 32)]/2

⇒ y = [6 ± 8]/2

⇒ y = 7 and y = - 1

Therefore, on solving, y = 3 ± √5, y = 7, and y = - 1.

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