Solve ) x ^ 2 + 15/16 = 2x | 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

$$x ^ { 2 } + \frac { 15 } { 16 } = 2 x$$

Solve for x

$x=\frac{3}{4}=0.75$
$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $2x$ from both sides.

$$x^{2}+\frac{15}{16}-2x=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$x^{2}-2x+\frac{15}{16}=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-2$ for $b$, and $\frac{15}{16}$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times \frac{15}{16}}}{2}$$

Square $-2$.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times \frac{15}{16}}}{2}$$

Multiply $-4$ times $\frac{15}{16}$.

$$x=\frac{-\left(-2\right)±\sqrt{4-\frac{15}{4}}}{2}$$

Add $4$ to $-\frac{15}{4}$.

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

Take the square root of $\frac{1}{4}$.

$$x=\frac{-\left(-2\right)±\frac{1}{2}}{2}$$

The opposite of $-2$ is $2$.

$$x=\frac{2±\frac{1}{2}}{2}$$

Now solve the equation $x=\frac{2±\frac{1}{2}}{2}$ when $±$ is plus. Add $2$ to $\frac{1}{2}$.

$$x=\frac{\frac{5}{2}}{2}$$

Divide $\frac{5}{2}$ by $2$.

$$x=\frac{5}{4}$$

Now solve the equation $x=\frac{2±\frac{1}{2}}{2}$ when $±$ is minus. Subtract $\frac{1}{2}$ from $2$.

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

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

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

The equation is now solved.

$$x=\frac{5}{4}$$ $$x=\frac{3}{4}$$

Steps for Completing the Square

Subtract $2x$ from both sides.

$$x^{2}+\frac{15}{16}-2x=0$$

Subtract $\frac{15}{16}$ from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-2x=-\frac{15}{16}$$

Divide $-2$, the coefficient of the $x$ term, by $2$ to get $-1$. Then add the square of $-1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=-\frac{15}{16}+1$$

Add $-\frac{15}{16}$ to $1$.

$$x^{2}-2x+1=\frac{1}{16}$$

Factor $x^{2}-2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-1\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

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

Add $1$ to both sides of the equation.

$$x=\frac{5}{4}$$ $$x=\frac{3}{4}$$