What is 5x^2+3x>= 3+2x^2 ?

The solution to 5x^2+3x>= 3+2x^2 is x = (sqrt(5)-1)/2

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5x^2+3x>= 3+2x^2

5 x^{2} + 3 x \geq 3 + 2 x^{2}

x \leq \frac{- 5 - 1}{2} or x \geq \frac{5 - 1}{2}

Interval Notation

(- \infty, \frac{- 5 - 1}{2}] \cup [\frac{5 - 1}{2}, \infty)

Decimal Notation

x \leq - 1.61803 \dots or x \geq 0.61803 \dots

Solution steps

5 x^{2} + 3 x \geq 3 + 2 x^{2}

Rewrite in standard form

x^{2} + x - 1 \geq 0

Complete the square

x^{2} + x - 1 : (x + \frac{1}{2})^{2} - \frac{5}{4}

(x + \frac{1}{2})^{2} - \frac{5}{4} \geq 0

Move

\frac{5}{4}

to the right side

(x + \frac{1}{2})^{2} \geq \frac{5}{4}

For

u^{n} \geq a

, if

n

is even then

x + \frac{1}{2} \leq - \frac{5}{4} or x + \frac{1}{2} \geq \frac{5}{4}

x + \frac{1}{2} \leq - \frac{5}{4} : x \leq \frac{- 5 - 1}{2}

x + \frac{1}{2} \geq \frac{5}{4} : x \geq \frac{5 - 1}{2}

Combine the intervals

x \leq \frac{- 5 - 1}{2} or x \geq \frac{5 - 1}{2}

\frac{2 + 3 i}{2 - 3 i}

3 (x - 9) = 18

\frac{y}{2} + \frac{5}{2} = \frac{y}{4}

lo g_{5} (\frac{1}{125}) = - 3

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

What is 5x^2+3x>= 3+2x^2 ?
The solution to 5x^2+3x>= 3+2x^2 is x<= (-sqrt(5)-1)/2 \lor x>= (sqrt(5)-1)/2