What are the solutions to the equation (x-3)(x+3)=4,x^2-9=4x^2-36 ?

The solutions to the equation (x-3)(x+3)=4,x^2-9=4x^2-36 are x=sqrt(13),x=-sqrt(13)

Find the zeros of (x-3)(x+3)=4,x^2-9=4x^2-36

The zeros of (x-3)(x+3)=4,x^2-9=4x^2-36 are x=sqrt(13),x=-sqrt(13)

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(x - 3) (x + 3) = 4, x^{2} - 9 = 4 x^{2} - 36

x = 13, x = - 13

Decimal Notation

x = 3.60555 \dots, x = - 3.60555 \dots

Solution steps

(x - 3) (x + 3) = 4

Expand

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

x^{2} - 9 = 4

Move

4

to the left side

x^{2} - 13 = 0

Solve with the quadratic formula

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

0^{2} - 4 \cdot 1 \cdot (- 13) = 213

x_{1, 2} = \frac{- 0 \pm 2 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 2 13}{2 \cdot 1}, x_{2} = \frac{- 0 - 2 13}{2 \cdot 1}

x = \frac{- 0 + 2 13}{2 \cdot 1} : 13

x = \frac{- 0 - 2 13}{2 \cdot 1} : - 13

The solutions to the quadratic equation are:

x = 13, x = - 13

3 - 4 x^{2} = 0

20 x^{2} + 40 x + 17 = 0

9 g^{2} + 4 = 0

2 u^{2} + 7 u + 6 = 0

- 16 x^{2} + 144 = 0

What are the solutions to the equation (x-3)(x+3)=4,x^2-9=4x^2-36 ?
The solutions to the equation (x-3)(x+3)=4,x^2-9=4x^2-36 are x=sqrt(13),x=-sqrt(13)
Find the zeros of (x-3)(x+3)=4,x^2-9=4x^2-36
The zeros of (x-3)(x+3)=4,x^2-9=4x^2-36 are x=sqrt(13),x=-sqrt(13)