What are the solutions to the equation factor 27x^2+5=48x ?

The solutions to the equation factor 27x^2+5=48x are x= 1/9 ,x= 5/3

Find the zeros of factor 27x^2+5=48x

The zeros of factor 27x^2+5=48x are x= 1/9 ,x= 5/3

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27 x^{2} + 5 = 48 x

x = \frac{1}{9}, x = \frac{5}{3}

Decimal

x = 0.11111 \dots, x = 1.66666 \dots

Solution steps

27 x^{2} + 5 = 48 x

Move

48 x

to the left side

27 x^{2} + 5 - 48 x = 0

Write in the standard form

a x^{2} + b x + c = 0

27 x^{2} - 48 x + 5 = 0

Factor

27 x^{2} - 48 x + 5 : (9 x - 1) (3 x - 5)

(9 x - 1) (3 x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

9 x - 1 = 0 or 3 x - 5 = 0

Solve

9 x - 1 = 0 : x = \frac{1}{9}

Solve

3 x - 5 = 0 : x = \frac{5}{3}

The solutions to the quadratic equation are:

x = \frac{1}{9}, x = \frac{5}{3}

What are the solutions to the equation factor 27x^2+5=48x ?
The solutions to the equation factor 27x^2+5=48x are x= 1/9 ,x= 5/3
Find the zeros of factor 27x^2+5=48x
The zeros of factor 27x^2+5=48x are x= 1/9 ,x= 5/3