Solve (x)^(2)-(9x)/(2)+2=0 | 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

$${ x }^{ 2 } - \frac{ 9x }{ 2 } +2=0$$

Solve for x

$x=\frac{1}{2}=0.5$
$x=4$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Multiply both sides of the equation by $2$.

$$2x^{2}-9x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2x^{2}+ax+bx+4$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-9$$ $$ab=2\times 4=8$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $8$.

$$-1,-8$$ $$-2,-4$$

Calculate the sum for each pair.

$$-1-8=-9$$ $$-2-4=-6$$

The solution is the pair that gives sum $-9$.

$$a=-8$$ $$b=-1$$

Rewrite $2x^{2}-9x+4$ as $\left(2x^{2}-8x\right)+\left(-x+4\right)$.

$$\left(2x^{2}-8x\right)+\left(-x+4\right)$$

Factor out $2x$ in the first and $-1$ in the second group.

$$2x\left(x-4\right)-\left(x-4\right)$$

Factor out common term $x-4$ by using distributive property.

$$\left(x-4\right)\left(2x-1\right)$$

To find equation solutions, solve $x-4=0$ and $2x-1=0$.

$$x=4$$ $$x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by $2$.

$$2x^{2}-9x+4=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-9$ for $b$, and $4$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 4}}{2\times 2}$$

Square $-9$.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 4}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 4}}{2\times 2}$$

Multiply $-8$ times $4$.

$$x=\frac{-\left(-9\right)±\sqrt{81-32}}{2\times 2}$$

Add $81$ to $-32$.

$$x=\frac{-\left(-9\right)±\sqrt{49}}{2\times 2}$$

Take the square root of $49$.

$$x=\frac{-\left(-9\right)±7}{2\times 2}$$

The opposite of $-9$ is $9$.

$$x=\frac{9±7}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{9±7}{4}$$

Now solve the equation $x=\frac{9±7}{4}$ when $±$ is plus. Add $9$ to $7$.

$$x=\frac{16}{4}$$

Divide $16$ by $4$.

$$x=4$$

Now solve the equation $x=\frac{9±7}{4}$ when $±$ is minus. Subtract $7$ from $9$.

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

Reduce the fraction $\frac{2}{4}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=4$$ $$x=\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by $2$.

$$2x^{2}-9x+4=0$$

Subtract $4$ from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-9x=-4$$

Divide both sides by $2$.

$$\frac{2x^{2}-9x}{2}=-\frac{4}{2}$$

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

$$x^{2}-\frac{9}{2}x=-\frac{4}{2}$$

Divide $-4$ by $2$.

$$x^{2}-\frac{9}{2}x=-2$$

Divide $-\frac{9}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{9}{4}$. Then add the square of $-\frac{9}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=-2+\left(-\frac{9}{4}\right)^{2}$$

Square $-\frac{9}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-2+\frac{81}{16}$$

Add $-2$ to $\frac{81}{16}$.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{49}{16}$$

Factor $x^{2}-\frac{9}{2}x+\frac{81}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{7}{4}$$ $$x-\frac{9}{4}=-\frac{7}{4}$$

Add $\frac{9}{4}$ to both sides of the equation.

$$x=4$$ $$x=\frac{1}{2}$$