What are the solutions to the equation 4.5(1-x)^2=4x^2 ?

The solutions to the equation 4.5(1-x)^2=4x^2 are x=3(3+2sqrt(2)),x=3(3-2sqrt(2))

Find the zeros of 4.5(1-x)^2=4x^2

The zeros of 4.5(1-x)^2=4x^2 are x=3(3+2sqrt(2)),x=3(3-2sqrt(2))

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4.5 (1 - x)^{2} = 4 x^{2}

x = 3 (3 + 22), x = 3 (3 - 22)

Decimal Notation

x = 17.48528 \dots, x = 0.51471 \dots

Solution steps

4.5 (1 - x)^{2} = 4 x^{2}

Multiply both sides by

10

45 (1 - x)^{2} = 40 x^{2}

Expand

45 (1 - x)^{2} : 45 - 90 x + 45 x^{2}

45 - 90 x + 45 x^{2} = 40 x^{2}

Move

40 x^{2}

to the left side

45 - 90 x + 5 x^{2} = 0

Write in the standard form

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

5 x^{2} - 90 x + 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 90 ) \pm ( - 90 ) ^{2} - 4 \cdot 5 \cdot 45}{2 \cdot 5}

(- 90)^{2} - 4 \cdot 5 \cdot 45 = 602

x_{1, 2} = \frac{- ( - 90 ) \pm 60 2}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 90 ) + 60 2}{2 \cdot 5}, x_{2} = \frac{- ( - 90 ) - 60 2}{2 \cdot 5}

x = \frac{- ( - 90 ) + 60 2}{2 \cdot 5} : 3 (3 + 22)

x = \frac{- ( - 90 ) - 60 2}{2 \cdot 5} : 3 (3 - 22)

The solutions to the quadratic equation are:

x = 3 (3 + 22), x = 3 (3 - 22)

(0.00078 x)^{2}

x + 6 \leq 9

∣ 4 Y - 4 ∣ + 4 \leq 16

lo g_{3} (\frac{1}{3})

6 x^{2} - 3 x - 5 = 0

What are the solutions to the equation 4.5(1-x)^2=4x^2 ?
The solutions to the equation 4.5(1-x)^2=4x^2 are x=3(3+2sqrt(2)),x=3(3-2sqrt(2))
Find the zeros of 4.5(1-x)^2=4x^2
The zeros of 4.5(1-x)^2=4x^2 are x=3(3+2sqrt(2)),x=3(3-2sqrt(2))