What are the solutions to the equation (7x-28)(28x-7)=0 ?

The solutions to the equation (7x-28)(28x-7)=0 are x=4,x= 1/4

Find the zeros of (7x-28)(28x-7)=0

The zeros of (7x-28)(28x-7)=0 are x=4,x= 1/4

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(7 x - 28) (28 x - 7) = 0

x = 4, x = \frac{1}{4}

Solution steps

(7 x - 28) (28 x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

7 x - 28 = 0 or 28 x - 7 = 0

Solve

7 x - 28 = 0 : x = 4

Solve

28 x - 7 = 0 : x = \frac{1}{4}

The solutions to the quadratic equation are:

x = 4, x = \frac{1}{4}

- 16 t^{2} + 48 t + 64 = 0

lo g_{\frac{1}{3}} (x) > lo g_{x} (3 - \frac{5}{2})

3 x^{2} + 18 x + 24 \geq 0

x^{2} - 2 x = 6

p = 3.51 - 50 + (48) e^{- 0.05 (0.25)}

What are the solutions to the equation (7x-28)(28x-7)=0 ?
The solutions to the equation (7x-28)(28x-7)=0 are x=4,x= 1/4
Find the zeros of (7x-28)(28x-7)=0
The zeros of (7x-28)(28x-7)=0 are x=4,x= 1/4