What are the solutions to the equation ((4x^2-1))/3 =x(10x-9) ?

The solutions to the equation ((4x^2-1))/3 =x(10x-9) are x= 1/26 ,x=1

Find the zeros of ((4x^2-1))/3 =x(10x-9)

The zeros of ((4x^2-1))/3 =x(10x-9) are x= 1/26 ,x=1

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\frac{( 4 x ^{2} - 1 )}{3} = x (10 x - 9)

x = \frac{1}{26}, x = 1

Decimal Notation

x = 0.03846 \dots, x = 1

Solution steps

\frac{( 4 x ^{2} - 1 )}{3} = x (10 x - 9)

Multiply both sides by

3

4 x^{2} - 1 = 3 x (10 x - 9)

Expand

3 x (10 x - 9) : 30 x^{2} - 27 x

4 x^{2} - 1 = 30 x^{2} - 27 x

Move

27 x

to the left side

4 x^{2} - 1 + 27 x = 30 x^{2}

Move

30 x^{2}

to the left side

- 26 x^{2} + 27 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 ( - 26 ) ( - 1 )}{2 ( - 26 )}

2 7^{2} - 4 (- 26) (- 1) = 25

x_{1, 2} = \frac{- 27 \pm 25}{2 ( - 26 )}

Separate the solutions

x_{1} = \frac{- 27 + 25}{2 ( - 26 )}, x_{2} = \frac{- 27 - 25}{2 ( - 26 )}

x = \frac{- 27 + 25}{2 ( - 26 )} : \frac{1}{26}

x = \frac{- 27 - 25}{2 ( - 26 )} : 1

The solutions to the quadratic equation are:

x = \frac{1}{26}, x = 1

a^{6} - 64

∣ 1 - 5 x ∣ + 2 \geq 9

\frac{13 !}{5 !}

\frac{x}{4} = \frac{5}{6}

What are the solutions to the equation ((4x^2-1))/3 =x(10x-9) ?
The solutions to the equation ((4x^2-1))/3 =x(10x-9) are x= 1/26 ,x=1
Find the zeros of ((4x^2-1))/3 =x(10x-9)
The zeros of ((4x^2-1))/3 =x(10x-9) are x= 1/26 ,x=1