What are the solutions to the equation x^2=(x-9)^2+(x-8)^2 ?

The solutions to the equation x^2=(x-9)^2+(x-8)^2 are x=29,x=5

Find the zeros of x^2=(x-9)^2+(x-8)^2

The zeros of x^2=(x-9)^2+(x-8)^2 are x=29,x=5

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x^{2} = (x - 9)^{2} + (x - 8)^{2}

x = 29, x = 5

Solution steps

x^{2} = (x - 9)^{2} + (x - 8)^{2}

Expand

(x - 9)^{2} + (x - 8)^{2} : 2 x^{2} - 34 x + 145

x^{2} = 2 x^{2} - 34 x + 145

Switch sides

2 x^{2} - 34 x + 145 = x^{2}

Move

x^{2}

to the left side

x^{2} - 34 x + 145 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 34 ) \pm ( - 34 ) ^{2} - 4 \cdot 1 \cdot 145}{2 \cdot 1}

(- 34)^{2} - 4 \cdot 1 \cdot 145 = 24

x_{1, 2} = \frac{- ( - 34 ) \pm 24}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 34 ) + 24}{2 \cdot 1}, x_{2} = \frac{- ( - 34 ) - 24}{2 \cdot 1}

x = \frac{- ( - 34 ) + 24}{2 \cdot 1} : 29

x = \frac{- ( - 34 ) - 24}{2 \cdot 1} : 5

The solutions to the quadratic equation are:

x = 29, x = 5

x^{2} = 3^{2} + 5^{2}

7 x^{2} - 25 x = 0

t, d = V t + \frac{1}{2} a t^{2}

2 x^{2} - 57 x + 216 = 0

x^{2} + 4 x + 4

What are the solutions to the equation x^2=(x-9)^2+(x-8)^2 ?
The solutions to the equation x^2=(x-9)^2+(x-8)^2 are x=29,x=5
Find the zeros of x^2=(x-9)^2+(x-8)^2
The zeros of x^2=(x-9)^2+(x-8)^2 are x=29,x=5