Solve b x ^ 2 = - 2.5x + 25/16 | 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

$$x ^ { 2 } = - 2.5 x + \frac { 25 } { 16 }$$

Solve for x

$x=\frac{5\sqrt{2}-5}{4}\approx 0.517766953$
$x=\frac{-5\sqrt{2}-5}{4}\approx -3.017766953$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $2.5x$ to both sides.

$$x^{2}+2.5x=\frac{25}{16}$$

Subtract $\frac{25}{16}$ from both sides.

$$x^{2}+2.5x-\frac{25}{16}=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2.5$ for $b$, and $-\frac{25}{16}$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2.5±\sqrt{2.5^{2}-4\left(-\frac{25}{16}\right)}}{2}$$

Square $2.5$ by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-2.5±\sqrt{6.25-4\left(-\frac{25}{16}\right)}}{2}$$

Multiply $-4$ times $-\frac{25}{16}$.

$$x=\frac{-2.5±\sqrt{\frac{25+25}{4}}}{2}$$

Add $6.25$ to $\frac{25}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-2.5±\sqrt{\frac{25}{2}}}{2}$$

Take the square root of $\frac{25}{2}$.

$$x=\frac{-2.5±\frac{5\sqrt{2}}{2}}{2}$$

Now solve the equation $x=\frac{-2.5±\frac{5\sqrt{2}}{2}}{2}$ when $±$ is plus. Add $-2.5$ to $\frac{5\sqrt{2}}{2}$.

$$x=\frac{5\sqrt{2}-5}{2\times 2}$$

Divide $\frac{-5+5\sqrt{2}}{2}$ by $2$.

$$x=\frac{5\sqrt{2}-5}{4}$$

Now solve the equation $x=\frac{-2.5±\frac{5\sqrt{2}}{2}}{2}$ when $±$ is minus. Subtract $\frac{5\sqrt{2}}{2}$ from $-2.5$.

$$x=\frac{-5\sqrt{2}-5}{2\times 2}$$

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

$$x=\frac{-5\sqrt{2}-5}{4}$$

The equation is now solved.

$$x=\frac{5\sqrt{2}-5}{4}$$ $$x=\frac{-5\sqrt{2}-5}{4}$$

Steps for Completing the Square

Add $2.5x$ to both sides.

$$x^{2}+2.5x=\frac{25}{16}$$

Divide $2.5$, the coefficient of the $x$ term, by $2$ to get $1.25$. Then add the square of $1.25$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2.5x+1.25^{2}=\frac{25}{16}+1.25^{2}$$

Square $1.25$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+2.5x+1.5625=\frac{25+25}{16}$$

Add $\frac{25}{16}$ to $1.5625$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+2.5x+1.5625=\frac{25}{8}$$

Factor $x^{2}+2.5x+1.5625$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+1.25\right)^{2}=\frac{25}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1.25\right)^{2}}=\sqrt{\frac{25}{8}}$$

Simplify.

$$x+1.25=\frac{5\sqrt{2}}{4}$$ $$x+1.25=-\frac{5\sqrt{2}}{4}$$

Subtract $1.25$ from both sides of the equation.

$$x=\frac{5\sqrt{2}-5}{4}$$ $$x=\frac{-5\sqrt{2}-5}{4}$$