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

The solutions to the equation (5sqrt(x^2+x))^2=(3x+3)^2 are x=-1,x= 9/16

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

The zeros of (5sqrt(x^2+x))^2=(3x+3)^2 are x=-1,x= 9/16

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

x = - 1, x = \frac{9}{16}

Decimal Notation

x = - 1, x = 0.5625

Solution steps

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

Expand

(5 x^{2} + x)^{2} : 25 x^{2} + 25 x

Expand

(3 x + 3)^{2} : 9 x^{2} + 18 x + 9

25 x^{2} + 25 x = 9 x^{2} + 18 x + 9

Switch sides

9 x^{2} + 18 x + 9 = 25 x^{2} + 25 x

Move

25 x

to the left side

9 x^{2} - 7 x + 9 = 25 x^{2}

Move

25 x^{2}

to the left side

- 16 x^{2} - 7 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 16 ) \cdot 9}{2 ( - 16 )}

(- 7)^{2} - 4 (- 16) \cdot 9 = 25

x_{1, 2} = \frac{- ( - 7 ) \pm 25}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 25}{2 ( - 16 )}, x_{2} = \frac{- ( - 7 ) - 25}{2 ( - 16 )}

x = \frac{- ( - 7 ) + 25}{2 ( - 16 )} : - 1

x = \frac{- ( - 7 ) - 25}{2 ( - 16 )} : \frac{9}{16}

The solutions to the quadratic equation are:

x = - 1, x = \frac{9}{16}

x, x^{2} + y^{2} - 6 x - 8 y = - 25

x - x^{2}

(x - 1) (x - 3) = - 1

6 a^{2} - 4 a - 22 = 0

x, 3 x^{2} + 2 x y + y^{2} = 1

What are the solutions to the equation (5sqrt(x^2+x))^2=(3x+3)^2 ?
The solutions to the equation (5sqrt(x^2+x))^2=(3x+3)^2 are x=-1,x= 9/16
Find the zeros of (5sqrt(x^2+x))^2=(3x+3)^2
The zeros of (5sqrt(x^2+x))^2=(3x+3)^2 are x=-1,x= 9/16