What are the solutions to the equation (2x-1)^2+(3x-2)^2=5 ?

The solutions to the equation (2x-1)^2+(3x-2)^2=5 are x= 16/13 ,x=0

Find the zeros of (2x-1)^2+(3x-2)^2=5

The zeros of (2x-1)^2+(3x-2)^2=5 are x= 16/13 ,x=0

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(2 x - 1)^{2} + (3 x - 2)^{2} = 5

x = \frac{16}{13}, x = 0

Solution steps

(2 x - 1)^{2} + (3 x - 2)^{2} = 5

Expand

(2 x - 1)^{2} + (3 x - 2)^{2} : 13 x^{2} - 16 x + 5

13 x^{2} - 16 x + 5 = 5

Move

5

to the left side

13 x^{2} - 16 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 13 \cdot 0}{2 \cdot 13}

(- 16)^{2} - 4 \cdot 13 \cdot 0 = 16

x_{1, 2} = \frac{- ( - 16 ) \pm 16}{2 \cdot 13}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 16}{2 \cdot 13}, x_{2} = \frac{- ( - 16 ) - 16}{2 \cdot 13}

x = \frac{- ( - 16 ) + 16}{2 \cdot 13} : \frac{16}{13}

x = \frac{- ( - 16 ) - 16}{2 \cdot 13} : 0

The solutions to the quadratic equation are:

x = \frac{16}{13}, x = 0

2 (x - 4)^{2} = 200

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

2 x^{2} - 4 = 3 x

(3 x - 1) (2 x + 3) = 8

2 m^{2} + m = 1

What are the solutions to the equation (2x-1)^2+(3x-2)^2=5 ?
The solutions to the equation (2x-1)^2+(3x-2)^2=5 are x= 16/13 ,x=0
Find the zeros of (2x-1)^2+(3x-2)^2=5
The zeros of (2x-1)^2+(3x-2)^2=5 are x= 16/13 ,x=0