Solve for x -3/2x^2=x+1

$- \frac{3}{2} x^{2} = x + 1$

Simplify $- \frac{3}{2} x^{2}$ .

Rewrite.

$0 + 0 - \frac{3}{2} x^{2} = x + 1$

Simplify by adding zeros.

$- \frac{3}{2} x^{2} = x + 1$

Combine $x^{2}$ and $\frac{3}{2}$ .

$- \frac{x^{2} \cdot 3}{2} = x + 1$

Move $3$ to the left of $x^{2}$ .

$- \frac{3 x^{2}}{2} = x + 1$

Subtract $x$ from both sides of the equation.

$- \frac{3 x^{2}}{2} - x = 1$

Subtract $1$ from both sides of the equation.

$- \frac{3 x^{2}}{2} - x - 1 = 0$

Multiply through by the least common denominator $2$ , then simplify.

Apply the distributive property.

$2 (- \frac{3 x^{2}}{2}) + 2 (- x) + 2 \cdot - 1 = 0$

Simplify.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3 x^{2}}{2}$ into the numerator.

$2 (\frac{- 3 x^{2}}{2}) + 2 (- x) + 2 \cdot - 1 = 0$

Cancel the common factor.

$2 (\frac{- 3 x^{2}}{2}) + 2 (- x) + 2 \cdot - 1 = 0$

Rewrite the expression.

$- 3 x^{2} + 2 (- x) + 2 \cdot - 1 = 0$

Multiply $- 1$ by $2$ .

$- 3 x^{2} - 2 x + 2 \cdot - 1 = 0$

Multiply $2$ by $- 1$ .

$- 3 x^{2} - 2 x - 2 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 3$ , $b = - 2$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (- 3 \cdot - 2)}}{2 \cdot - 3}$

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 3 \cdot - 2}}{2 \cdot - 3}$

Multiply $- 4$ by $- 3$ .

$x = \frac{2 \pm \sqrt{4 + 12 \cdot - 2}}{2 \cdot - 3}$

Multiply $12$ by $- 2$ .

$x = \frac{2 \pm \sqrt{4 - 24}}{2 \cdot - 3}$

Subtract $24$ from $4$ .

$x = \frac{2 \pm \sqrt{- 20}}{2 \cdot - 3}$

Rewrite $- 20$ as $- 1 (20)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 20}}{2 \cdot - 3}$

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{20}}{2 \cdot - 3}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{20}}{2 \cdot - 3}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (5)}}{2 \cdot - 3}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 5}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{5})}{2 \cdot - 3}$

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{5}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{2 \pm 2 i \sqrt{5}}{- 6}$

Simplify $\frac{2 \pm 2 i \sqrt{5}}{- 6}$ .

$x = \frac{1 \pm i \sqrt{5}}{- 3}$

Move the negative in front of the fraction.

$x = - \frac{1 \pm i \sqrt{5}}{3}$

The final answer is the combination of both solutions.

$x = - \frac{1 + i \sqrt{5}}{3}, - \frac{1 - i \sqrt{5}}{3}$

Solve for x -3/2x^2=x+1

Related Questions

Support LCMGCF.com by disabling your adblocker.