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

The solutions to the equation (4x-2)^2-(x-2)^2=9 are x=(2+sqrt(19))/5 ,x=(2-sqrt(19))/5

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

The zeros of (4x-2)^2-(x-2)^2=9 are x=(2+sqrt(19))/5 ,x=(2-sqrt(19))/5

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

x = \frac{2 + 19}{5}, x = \frac{2 - 19}{5}

Solution steps

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

Expand

(4 x - 2)^{2} - (x - 2)^{2} : 15 x^{2} - 12 x

15 x^{2} - 12 x = 9

Move

9

to the left side

15 x^{2} - 12 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 15 ( - 9 )}{2 \cdot 15}

(- 12)^{2} - 4 \cdot 15 (- 9) = 619

x_{1, 2} = \frac{- ( - 12 ) \pm 6 19}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 6 19}{2 \cdot 15}, x_{2} = \frac{- ( - 12 ) - 6 19}{2 \cdot 15}

x = \frac{- ( - 12 ) + 6 19}{2 \cdot 15} : \frac{2 + 19}{5}

x = \frac{- ( - 12 ) - 6 19}{2 \cdot 15} : \frac{2 - 19}{5}

The solutions to the quadratic equation are:

x = \frac{2 + 19}{5}, x = \frac{2 - 19}{5}

∣ 9 - 2 x ∣ = 3

(x - 1) (x + 1) + (x - 1) (x + 2)

x y - x - y + 1

0 = ∣ x + 5 ∣ - 9

ln (\frac{1}{7})

What are the solutions to the equation (4x-2)^2-(x-2)^2=9 ?
The solutions to the equation (4x-2)^2-(x-2)^2=9 are x=(2+sqrt(19))/5 ,x=(2-sqrt(19))/5
Find the zeros of (4x-2)^2-(x-2)^2=9
The zeros of (4x-2)^2-(x-2)^2=9 are x=(2+sqrt(19))/5 ,x=(2-sqrt(19))/5