Solve 25-(2)^(2)x=x(x\times1) | 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

$$25- { 2 }^{ 2 } x=x(x \times 1)$$

Solve for x (complex solution)

$x=\sqrt{29}-2\approx 3.385164807$
$x=-\left(\sqrt{29}+2\right)\approx -7.385164807$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$25-2^{2}x=x^{2}\times 1$$

Calculate $2$ to the power of $2$ and get $4$.

$$25-4x=x^{2}\times 1$$

Subtract $x^{2}\times 1$ from both sides.

$$25-4x-x^{2}\times 1=0$$

Reorder the terms.

$$-4x+25-x^{2}=0$$

Reorder the terms.

$$-x^{2}-4x+25=0$$

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

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 25}}{2\left(-1\right)}$$

Multiply $4$ times $25$.

$$x=\frac{-\left(-4\right)±\sqrt{16+100}}{2\left(-1\right)}$$

Add $16$ to $100$.

$$x=\frac{-\left(-4\right)±\sqrt{116}}{2\left(-1\right)}$$

Take the square root of $116$.

$$x=\frac{-\left(-4\right)±2\sqrt{29}}{2\left(-1\right)}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±2\sqrt{29}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{4±2\sqrt{29}}{-2}$$

Now solve the equation $x=\frac{4±2\sqrt{29}}{-2}$ when $±$ is plus. Add $4$ to $2\sqrt{29}$.

$$x=\frac{2\sqrt{29}+4}{-2}$$

Divide $4+2\sqrt{29}$ by $-2$.

$$x=-\left(\sqrt{29}+2\right)$$

Now solve the equation $x=\frac{4±2\sqrt{29}}{-2}$ when $±$ is minus. Subtract $2\sqrt{29}$ from $4$.

$$x=\frac{4-2\sqrt{29}}{-2}$$

Divide $4-2\sqrt{29}$ by $-2$.

$$x=\sqrt{29}-2$$

The equation is now solved.

$$x=-\left(\sqrt{29}+2\right)$$ $$x=\sqrt{29}-2$$

Steps for Completing the Square

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

$$25-2^{2}x=x^{2}\times 1$$

Calculate $2$ to the power of $2$ and get $4$.

$$25-4x=x^{2}\times 1$$

Subtract $x^{2}\times 1$ from both sides.

$$25-4x-x^{2}\times 1=0$$

Reorder the terms.

$$-4x+25-x^{2}=0$$

Subtract $25$ from both sides. Anything subtracted from zero gives its negation.

$$-4x-x^{2}=-25$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}-4x=-25$$

Divide both sides by $-1$.

$$\frac{-x^{2}-4x}{-1}=-\frac{25}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{25}{-1}$$

Divide $-4$ by $-1$.

$$x^{2}+4x=-\frac{25}{-1}$$

Divide $-25$ by $-1$.

$$x^{2}+4x=25$$

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

$$x^{2}+4x+2^{2}=25+2^{2}$$

Square $2$.

$$x^{2}+4x+4=25+4$$

Add $25$ to $4$.

$$x^{2}+4x+4=29$$

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

$$\left(x+2\right)^{2}=29$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{29}$$

Simplify.

$$x+2=\sqrt{29}$$ $$x+2=-\sqrt{29}$$

Subtract $2$ from both sides of the equation.

$$x=\sqrt{29}-2$$ $$x=-\sqrt{29}-2$$

Solve for x

$x=\sqrt{29}-2\approx 3.385164807$
$x=-\sqrt{29}-2\approx -7.385164807$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$25-2^{2}x=x^{2}\times 1$$

Calculate $2$ to the power of $2$ and get $4$.

$$25-4x=x^{2}\times 1$$

Subtract $x^{2}\times 1$ from both sides.

$$25-4x-x^{2}\times 1=0$$

Reorder the terms.

$$-4x+25-x^{2}=0$$

Reorder the terms.

$$-x^{2}-4x+25=0$$

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

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 25}}{2\left(-1\right)}$$

Multiply $4$ times $25$.

$$x=\frac{-\left(-4\right)±\sqrt{16+100}}{2\left(-1\right)}$$

Add $16$ to $100$.

$$x=\frac{-\left(-4\right)±\sqrt{116}}{2\left(-1\right)}$$

Take the square root of $116$.

$$x=\frac{-\left(-4\right)±2\sqrt{29}}{2\left(-1\right)}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±2\sqrt{29}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{4±2\sqrt{29}}{-2}$$

Now solve the equation $x=\frac{4±2\sqrt{29}}{-2}$ when $±$ is plus. Add $4$ to $2\sqrt{29}$.

$$x=\frac{2\sqrt{29}+4}{-2}$$

Divide $4+2\sqrt{29}$ by $-2$.

$$x=-\left(\sqrt{29}+2\right)$$

Now solve the equation $x=\frac{4±2\sqrt{29}}{-2}$ when $±$ is minus. Subtract $2\sqrt{29}$ from $4$.

$$x=\frac{4-2\sqrt{29}}{-2}$$

Divide $4-2\sqrt{29}$ by $-2$.

$$x=\sqrt{29}-2$$

The equation is now solved.

$$x=-\left(\sqrt{29}+2\right)$$ $$x=\sqrt{29}-2$$

Steps for Completing the Square

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

$$25-2^{2}x=x^{2}\times 1$$

Calculate $2$ to the power of $2$ and get $4$.

$$25-4x=x^{2}\times 1$$

Subtract $x^{2}\times 1$ from both sides.

$$25-4x-x^{2}\times 1=0$$

Reorder the terms.

$$-4x+25-x^{2}=0$$

Subtract $25$ from both sides. Anything subtracted from zero gives its negation.

$$-4x-x^{2}=-25$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}-4x=-25$$

Divide both sides by $-1$.

$$\frac{-x^{2}-4x}{-1}=-\frac{25}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{25}{-1}$$

Divide $-4$ by $-1$.

$$x^{2}+4x=-\frac{25}{-1}$$

Divide $-25$ by $-1$.

$$x^{2}+4x=25$$

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

$$x^{2}+4x+2^{2}=25+2^{2}$$

Square $2$.

$$x^{2}+4x+4=25+4$$

Add $25$ to $4$.

$$x^{2}+4x+4=29$$

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

$$\left(x+2\right)^{2}=29$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{29}$$

Simplify.

$$x+2=\sqrt{29}$$ $$x+2=-\sqrt{29}$$

Subtract $2$ from both sides of the equation.

$$x=\sqrt{29}-2$$ $$x=-\sqrt{29}-2$$