What are the solutions to the equation 5(x^2-3x-4)=(x^2-2x-3)*4 ?

The solutions to the equation 5(x^2-3x-4)=(x^2-2x-3)*4 are x=-1,x=8

Find the zeros of 5(x^2-3x-4)=(x^2-2x-3)*4

The zeros of 5(x^2-3x-4)=(x^2-2x-3)*4 are x=-1,x=8

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5 (x^{2} - 3 x - 4) = (x^{2} - 2 x - 3) \cdot 4

x = - 1, x = 8

Solution steps

5 (x^{2} - 3 x - 4) = (x^{2} - 2 x - 3) \cdot 4

Move

(x^{2} - 2 x - 3) \cdot 4

to the left side

5 (x^{2} - 3 x - 4) - (x^{2} - 2 x - 3) \cdot 4 = 0

Factor

5 (x^{2} - 3 x - 4) - (x^{2} - 2 x - 3) \cdot 4 : (x + 1) (x - 8)

(x + 1) (x - 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 8 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 8 = 0 : x = 8

The solutions to the quadratic equation are:

x = - 1, x = 8

x^{2} = - x + 12

(x^{2} - x - 3) lo g_{10} (4) = 3 lo g_{10} (\frac{1}{4})

k^{2} - 7 k + 50 = 3

2 y^{2} - 50 = 0

(4 - k) x^{2} - 2 k x = - 9

What are the solutions to the equation 5(x^2-3x-4)=(x^2-2x-3)*4 ?
The solutions to the equation 5(x^2-3x-4)=(x^2-2x-3)*4 are x=-1,x=8
Find the zeros of 5(x^2-3x-4)=(x^2-2x-3)*4
The zeros of 5(x^2-3x-4)=(x^2-2x-3)*4 are x=-1,x=8