What are the solutions to the equation (-18m^2)^2-4(1+m^2)(81m^2-49)=0 ?

The solutions to the equation (-18m^2)^2-4(1+m^2)(81m^2-49)=0 are m=-7/(4sqrt(2)),m= 7/(4sqrt(2))

Find the zeros of (-18m^2)^2-4(1+m^2)(81m^2-49)=0

The zeros of (-18m^2)^2-4(1+m^2)(81m^2-49)=0 are m=-7/(4sqrt(2)),m= 7/(4sqrt(2))

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(- 18 m^{2})^{2} - 4 (1 + m^{2}) (81 m^{2} - 49) = 0

m = - \frac{7}{4 2}, m = \frac{7}{4 2}

Decimal Notation

m = - 1.23743 \dots, m = 1.23743 \dots

Solution steps

(- 18 m^{2})^{2} - 4 (1 + m^{2}) (81 m^{2} - 49) = 0

Expand

(- 18 m^{2})^{2} - 4 (1 + m^{2}) (81 m^{2} - 49) : - 128 m^{2} + 196

- 128 m^{2} + 196 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 128 ) \cdot 196}{2 ( - 128 )}

0^{2} - 4 (- 128) \cdot 196 = 2242

m_{1, 2} = \frac{- 0 \pm 224 2}{2 ( - 128 )}

Separate the solutions

m_{1} = \frac{- 0 + 224 2}{2 ( - 128 )}, m_{2} = \frac{- 0 - 224 2}{2 ( - 128 )}

m = \frac{- 0 + 224 2}{2 ( - 128 )} : - \frac{7}{4 2}

m = \frac{- 0 - 224 2}{2 ( - 128 )} : \frac{7}{4 2}

The solutions to the quadratic equation are:

m = - \frac{7}{4 2}, m = \frac{7}{4 2}

p^{2} + p - 2 = 0

x^{4} - 8 x^{2} + 7

- 14 x + 3 = - 9 x - 12

x^{2} + 7 x + 6 \geq 0

(x - 1) (x + 4) = 36

What are the solutions to the equation (-18m^2)^2-4(1+m^2)(81m^2-49)=0 ?
The solutions to the equation (-18m^2)^2-4(1+m^2)(81m^2-49)=0 are m=-7/(4sqrt(2)),m= 7/(4sqrt(2))
Find the zeros of (-18m^2)^2-4(1+m^2)(81m^2-49)=0
The zeros of (-18m^2)^2-4(1+m^2)(81m^2-49)=0 are m=-7/(4sqrt(2)),m= 7/(4sqrt(2))