Solve x+ 2 x =14+7); Premfko)n^ 1000 +n^ -1000 =-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

$$x+\frac{2}{x}=14+7); Premfko)n^{1000}+n^{-1000}=-1$$

Solve for x

$x = \frac{\sqrt{433} + 21}{2} \approx 20.904326023$
$x=\frac{21-\sqrt{433}}{2}\approx 0.095673977$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$xx+2=x\times 14+x\times 7$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+2=x\times 14+x\times 7$$

Combine $x\times 14$ and $x\times 7$ to get $21x$.

$$x^{2}+2=21x$$

Subtract $21x$ from both sides.

$$x^{2}+2-21x=0$$

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.

$$x^{2}-21x+2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-21$ for $b$, and $2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 2}}{2}$$

Square $-21$.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 2}}{2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-21\right)±\sqrt{441-8}}{2}$$

Add $441$ to $-8$.

$$x=\frac{-\left(-21\right)±\sqrt{433}}{2}$$

The opposite of $-21$ is $21$.

$$x=\frac{21±\sqrt{433}}{2}$$

Now solve the equation $x=\frac{21±\sqrt{433}}{2}$ when $±$ is plus. Add $21$ to $\sqrt{433}$.

$$x=\frac{\sqrt{433}+21}{2}$$

Now solve the equation $x=\frac{21±\sqrt{433}}{2}$ when $±$ is minus. Subtract $\sqrt{433}$ from $21$.

$$x=\frac{21-\sqrt{433}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{433}+21}{2}$$ $$x=\frac{21-\sqrt{433}}{2}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$xx+2=x\times 14+x\times 7$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+2=x\times 14+x\times 7$$

Combine $x\times 14$ and $x\times 7$ to get $21x$.

$$x^{2}+2=21x$$

Subtract $21x$ from both sides.

$$x^{2}+2-21x=0$$

Subtract $2$ from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-21x=-2$$

Divide $-21$, the coefficient of the $x$ term, by $2$ to get $-\frac{21}{2}$. Then add the square of $-\frac{21}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-21x+\left(-\frac{21}{2}\right)^{2}=-2+\left(-\frac{21}{2}\right)^{2}$$

Square $-\frac{21}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-21x+\frac{441}{4}=-2+\frac{441}{4}$$

Add $-2$ to $\frac{441}{4}$.

$$x^{2}-21x+\frac{441}{4}=\frac{433}{4}$$

Factor $x^{2}-21x+\frac{441}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{21}{2}\right)^{2}=\frac{433}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{21}{2}\right)^{2}}=\sqrt{\frac{433}{4}}$$

Simplify.

$$x-\frac{21}{2}=\frac{\sqrt{433}}{2}$$ $$x-\frac{21}{2}=-\frac{\sqrt{433}}{2}$$

Add $\frac{21}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{433}+21}{2}$$ $$x=\frac{21-\sqrt{433}}{2}$$