What are the solutions to the equation x+9=4(x-1)^2 ?

The solutions to the equation x+9=4(x-1)^2 are x=(9+sqrt(161))/8 ,x=(9-sqrt(161))/8

Find the zeros of x+9=4(x-1)^2

The zeros of x+9=4(x-1)^2 are x=(9+sqrt(161))/8 ,x=(9-sqrt(161))/8

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x + 9 = 4 (x - 1)^{2}

x = \frac{9 + 161}{8}, x = \frac{9 - 161}{8}

Solution steps

x + 9 = 4 (x - 1)^{2}

Expand

4 (x - 1)^{2} : 4 x^{2} - 8 x + 4

x + 9 = 4 x^{2} - 8 x + 4

Switch sides

4 x^{2} - 8 x + 4 = x + 9

Move

9

to the left side

4 x^{2} - 8 x - 5 = x

Move

x

to the left side

4 x^{2} - 9 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

(- 9)^{2} - 4 \cdot 4 (- 5) = 161

x_{1, 2} = \frac{- ( - 9 ) \pm 161}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 161}{2 \cdot 4}, x_{2} = \frac{- ( - 9 ) - 161}{2 \cdot 4}

x = \frac{- ( - 9 ) + 161}{2 \cdot 4} : \frac{9 + 161}{8}

x = \frac{- ( - 9 ) - 161}{2 \cdot 4} : \frac{9 - 161}{8}

The solutions to the quadratic equation are:

x = \frac{9 + 161}{8}, x = \frac{9 - 161}{8}

14 \cdot 6

- 3 (x + 5) + 2 x = - 15 - x

(x - 4) (x + 1) (x - 6)^{2} < 0

27 a^{2} - 48

7^{2} - \frac{1}{2} - 6

What are the solutions to the equation x+9=4(x-1)^2 ?
The solutions to the equation x+9=4(x-1)^2 are x=(9+sqrt(161))/8 ,x=(9-sqrt(161))/8
Find the zeros of x+9=4(x-1)^2
The zeros of x+9=4(x-1)^2 are x=(9+sqrt(161))/8 ,x=(9-sqrt(161))/8