Solve 2x ^ 2 = 17/3 * x + 1 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$2x^{2}=\frac{17}{3}x+1$$

Solve for x

$x=-\frac{1}{6}\approx -0.166666667$
$x=3$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $\frac{17}{3}x$ from both sides.

$$2x^{2}-\frac{17}{3}x=1$$

Subtract $1$ from both sides.

$$2x^{2}-\frac{17}{3}x-1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-\frac{17}{3}$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\left(-\frac{17}{3}\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Square $-\frac{17}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\frac{289}{9}-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\frac{289}{9}-8\left(-1\right)}}{2\times 2}$$

Multiply $-8$ times $-1$.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\frac{289}{9}+8}}{2\times 2}$$

Add $\frac{289}{9}$ to $8$.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\frac{361}{9}}}{2\times 2}$$

Take the square root of $\frac{361}{9}$.

$$x=\frac{-\left(-\frac{17}{3}\right)±\frac{19}{3}}{2\times 2}$$

The opposite of $-\frac{17}{3}$ is $\frac{17}{3}$.

$$x=\frac{\frac{17}{3}±\frac{19}{3}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{\frac{17}{3}±\frac{19}{3}}{4}$$

Now solve the equation $x=\frac{\frac{17}{3}±\frac{19}{3}}{4}$ when $±$ is plus. Add $\frac{17}{3}$ to $\frac{19}{3}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{12}{4}$$

Divide $12$ by $4$.

$$x=3$$

Now solve the equation $x=\frac{\frac{17}{3}±\frac{19}{3}}{4}$ when $±$ is minus. Subtract $\frac{19}{3}$ from $\frac{17}{3}$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{2}{3}}{4}$$

Divide $-\frac{2}{3}$ by $4$.

$$x=-\frac{1}{6}$$

The equation is now solved.

$$x=3$$ $$x=-\frac{1}{6}$$

Steps for Completing the Square

Subtract $\frac{17}{3}x$ from both sides.

$$2x^{2}-\frac{17}{3}x=1$$

Divide both sides by $2$.

$$\frac{2x^{2}-\frac{17}{3}x}{2}=\frac{1}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}+\left(-\frac{\frac{17}{3}}{2}\right)x=\frac{1}{2}$$

Divide $-\frac{17}{3}$ by $2$.

$$x^{2}-\frac{17}{6}x=\frac{1}{2}$$

Divide $-\frac{17}{6}$, the coefficient of the $x$ term, by $2$ to get $-\frac{17}{12}$. Then add the square of $-\frac{17}{12}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{6}x+\left(-\frac{17}{12}\right)^{2}=\frac{1}{2}+\left(-\frac{17}{12}\right)^{2}$$

Square $-\frac{17}{12}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{6}x+\frac{289}{144}=\frac{1}{2}+\frac{289}{144}$$

Add $\frac{1}{2}$ to $\frac{289}{144}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{6}x+\frac{289}{144}=\frac{361}{144}$$

Factor $x^{2}-\frac{17}{6}x+\frac{289}{144}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{17}{12}\right)^{2}=\frac{361}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{12}\right)^{2}}=\sqrt{\frac{361}{144}}$$

Simplify.

$$x-\frac{17}{12}=\frac{19}{12}$$ $$x-\frac{17}{12}=-\frac{19}{12}$$

Add $\frac{17}{12}$ to both sides of the equation.

$$x=3$$ $$x=-\frac{1}{6}$$