What are the solutions to the equation 9x(4-x)=31 ?

The solutions to the equation 9x(4-x)=31 are x=(6-sqrt(5))/3 ,x=(6+sqrt(5))/3

Find the zeros of 9x(4-x)=31

The zeros of 9x(4-x)=31 are x=(6-sqrt(5))/3 ,x=(6+sqrt(5))/3

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9 x (4 - x) = 31

x = \frac{6 - 5}{3}, x = \frac{6 + 5}{3}

Decimal Notation

x = 1.25464 \dots, x = 2.74535 \dots

Solution steps

9 x (4 - x) = 31

Expand

9 x (4 - x) : 36 x - 9 x^{2}

36 x - 9 x^{2} = 31

Move

31

to the left side

36 x - 9 x^{2} - 31 = 0

Write in the standard form

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

- 9 x^{2} + 36 x - 31 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 ( - 9 ) ( - 31 )}{2 ( - 9 )}

3 6^{2} - 4 (- 9) (- 31) = 65

x_{1, 2} = \frac{- 36 \pm 6 5}{2 ( - 9 )}

Separate the solutions

x_{1} = \frac{- 36 + 6 5}{2 ( - 9 )}, x_{2} = \frac{- 36 - 6 5}{2 ( - 9 )}

x = \frac{- 36 + 6 5}{2 ( - 9 )} : \frac{6 - 5}{3}

x = \frac{- 36 - 6 5}{2 ( - 9 )} : \frac{6 + 5}{3}

The solutions to the quadratic equation are:

x = \frac{6 - 5}{3}, x = \frac{6 + 5}{3}

x^{2} = 1^{2} + (\frac{1}{2})^{2}

60 = 50 + 2 t - 0.0625 t^{2}

- 4 x^{2} + 2 x - 5 = 0

6 x^{2} + 6 x = 1

(x - 5)^{2} - 2 = 79

What are the solutions to the equation 9x(4-x)=31 ?
The solutions to the equation 9x(4-x)=31 are x=(6-sqrt(5))/3 ,x=(6+sqrt(5))/3
Find the zeros of 9x(4-x)=31
The zeros of 9x(4-x)=31 are x=(6-sqrt(5))/3 ,x=(6+sqrt(5))/3