What are the solutions to the equation 2(x+7)(25x^2-49)=0 ?

The solutions to the equation 2(x+7)(25x^2-49)=0 are x=-7,x=-7/5 ,x= 7/5

Find the zeros of 2(x+7)(25x^2-49)=0

The zeros of 2(x+7)(25x^2-49)=0 are x=-7,x=-7/5 ,x= 7/5

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2 (x + 7) (25 x^{2} - 49) = 0

x = - 7, x = - \frac{7}{5}, x = \frac{7}{5}

Decimal Notation

x = - 7, x = - 1.4, x = 1.4

Solution steps

2 (x + 7) (25 x^{2} - 49) = 0

Factor

2 (x + 7) (25 x^{2} - 49) : 2 (x + 7) (5 x + 7) (5 x - 7)

2 (x + 7) (5 x + 7) (5 x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 7 = 0 or 5 x + 7 = 0 or 5 x - 7 = 0

Solve

x + 7 = 0 : x = - 7

Solve

5 x + 7 = 0 : x = - \frac{7}{5}

Solve

5 x - 7 = 0 : x = \frac{7}{5}

The solutions are

x = - 7, x = - \frac{7}{5}, x = \frac{7}{5}

x^{9} + 3 x^{6} + 3 x^{3} - 16 x - 7 = 0

x^{3} - 3 x^{2} - 9 x + 11 = 0

x x^{2} + 6 x + 8 = 0

x^{3} - 28 = 100

x^{4} - 8 x^{2} + 1 = 0

What are the solutions to the equation 2(x+7)(25x^2-49)=0 ?
The solutions to the equation 2(x+7)(25x^2-49)=0 are x=-7,x=-7/5 ,x= 7/5
Find the zeros of 2(x+7)(25x^2-49)=0
The zeros of 2(x+7)(25x^2-49)=0 are x=-7,x=-7/5 ,x= 7/5