What are the solutions to the equation (3x-1)^2=(x+2)^2 ?

The solutions to the equation (3x-1)^2=(x+2)^2 are x= 3/2 ,x=-1/4

Find the zeros of (3x-1)^2=(x+2)^2

The zeros of (3x-1)^2=(x+2)^2 are x= 3/2 ,x=-1/4

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(3 x - 1)^{2} = (x + 2)^{2}

x = \frac{3}{2}, x = - \frac{1}{4}

Solution steps

(3 x - 1)^{2} = (x + 2)^{2}

Expand

(3 x - 1)^{2} : 9 x^{2} - 6 x + 1

Expand

(x + 2)^{2} : x^{2} + 4 x + 4

9 x^{2} - 6 x + 1 = x^{2} + 4 x + 4

Move

4

to the left side

9 x^{2} - 6 x - 3 = x^{2} + 4 x

Move

4 x

to the left side

9 x^{2} - 10 x - 3 = x^{2}

Move

x^{2}

to the left side

8 x^{2} - 10 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

(- 10)^{2} - 4 \cdot 8 (- 3) = 14

x_{1, 2} = \frac{- ( - 10 ) \pm 14}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 14}{2 \cdot 8}, x_{2} = \frac{- ( - 10 ) - 14}{2 \cdot 8}

x = \frac{- ( - 10 ) + 14}{2 \cdot 8} : \frac{3}{2}

x = \frac{- ( - 10 ) - 14}{2 \cdot 8} : - \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - \frac{1}{4}

\frac{2.0 5 ^{3}}{6} + 4 (2.05)

4 x - 7 = - 5 x^{2}

\frac{11}{64} \cdot \frac{180}{π}

∣ 5 x - 4 ∣ + 4 = 7

5 x = - 4 x + 36

What are the solutions to the equation (3x-1)^2=(x+2)^2 ?
The solutions to the equation (3x-1)^2=(x+2)^2 are x= 3/2 ,x=-1/4
Find the zeros of (3x-1)^2=(x+2)^2
The zeros of (3x-1)^2=(x+2)^2 are x= 3/2 ,x=-1/4