What are the solutions to the equation solvefor x,h(x)^2=x^2+5x ?

The solutions to the equation solvefor x,h(x)^2=x^2+5x are x=0,x=-5/(1-h);\quad h\ne 1

Find the zeros of solvefor x,h(x)^2=x^2+5x

The zeros of solvefor x,h(x)^2=x^2+5x are x=0,x=-5/(1-h);\quad h\ne 1

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solve for

x, h (x)^{2} = x^{2} + 5 x

x = 0, x = - \frac{5}{1 - h}; h \neq = 1

Solution steps

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

Switch sides

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

Move

h x^{2}

to the left side

x^{2} + 5 x - h x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

(1 - h) x^{2} + 5 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( 1 - h ) \cdot 0}{2 ( 1 - h )}

5^{2} - 4 (1 - h) \cdot 0 = 5

x_{1, 2} = \frac{- 5 \pm 5}{2 ( 1 - h )}; h \neq = 1

Separate the solutions

x_{1} = \frac{- 5 + 5}{2 ( 1 - h )}, x_{2} = \frac{- 5 - 5}{2 ( 1 - h )}

x = \frac{- 5 + 5}{2 ( 1 - h )} : 0

x = \frac{- 5 - 5}{2 ( 1 - h )} : - \frac{5}{1 - h}

The solutions to the quadratic equation are:

x = 0, x = - \frac{5}{1 - h}; h \neq = 1

\frac{( x - 6 ) ^{2}}{16} - \frac{( y - 4 ) ^{2}}{9} = 1

x^{2} - 3 y^{2} + 5 y = 7

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

x^{2} - 2 y^{2} = 8

25 y^{2} + 250 y - 16 x^{2} - 32 x + 209 = 0

What are the solutions to the equation solvefor x,h(x)^2=x^2+5x ?
The solutions to the equation solvefor x,h(x)^2=x^2+5x are x=0,x=-5/(1-h);\quad h\ne 1
Find the zeros of solvefor x,h(x)^2=x^2+5x
The zeros of solvefor x,h(x)^2=x^2+5x are x=0,x=-5/(1-h);\quad h\ne 1