What is the answer to c= 1/((sqrt(x^2+x-2))) ?

The answer to c= 1/((sqrt(x^2+x-2))) is x=(-c+sqrt(9c^2+4))/(2c),x=-(c+sqrt(9c^2+4))/(2c)

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c= 1/((sqrt(x^2+x-2)))

c = \frac{1}{( x ^{2} + x - 2 )}

x = \frac{- c + 9 c ^{2} + 4}{2 c}, x = - \frac{c + 9 c ^{2} + 4}{2 c}

Solution steps

c = \frac{1}{( x ^{2} + x - 2 )}

Multiply both sides by

x^{2} + x - 2

c x^{2} + x - 2 = \frac{1}{x ^{2} + x - 2} x^{2} + x - 2

Simplify

c x^{2} + x - 2 = 1

Divide both sides by

c

x^{2} + x - 2 = \frac{1}{c}

Square both sides:

x^{2} + x - 2 = \frac{1}{c ^{2}}

x^{2} + x - 2 = \frac{1}{c ^{2}}

Solve

x^{2} + x - 2 = \frac{1}{c ^{2}} : x = \frac{- c + 9 c ^{2} + 4}{2 c}, x = - \frac{c + 9 c ^{2} + 4}{2 c}

x = \frac{- c + 9 c ^{2} + 4}{2 c}, x = - \frac{c + 9 c ^{2} + 4}{2 c}

Verify Solutions:

x = \frac{- c + 9 c ^{2} + 4}{2 c}

True

, x = - \frac{c + 9 c ^{2} + 4}{2 c}

True

The solutions are

x = \frac{- c + 9 c ^{2} + 4}{2 c}, x = - \frac{c + 9 c ^{2} + 4}{2 c}

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

(\frac{x - 1}{x})^{\frac{1}{2}} + (\frac{1 - 1}{x})^{\frac{1}{2}} = x

x + 1 + 2 x + 3 = 5

2 x - 3 = - 1

x, (a) f = - 6 - 3 x

What is the answer to c= 1/((sqrt(x^2+x-2))) ?
The answer to c= 1/((sqrt(x^2+x-2))) is x=(-c+sqrt(9c^2+4))/(2c),x=-(c+sqrt(9c^2+4))/(2c)