How to solve x+1=(x)^(2)-6x+9

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve x+1=(x)^(2)-6x+9 | Equatix

Question

$$x+1= { x }^{ 2 } -6x+9$$

Solve for x

$x = \frac{\sqrt{17} + 7}{2} \approx 5.561552813$
$x = \frac{7 - \sqrt{17}}{2} \approx 1.438447187$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $x^{2}$ from both sides.

$$x+1-x^{2}=-6x+9$$

Add $6x$ to both sides.

$$x+1-x^{2}+6x=9$$

Combine $x$ and $6x$ to get $7x$.

$$7x+1-x^{2}=9$$

Subtract $9$ from both sides.

$$7x+1-x^{2}-9=0$$

Subtract $9$ from $1$ to get $-8$.

$$7x-8-x^{2}=0$$

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.

$$-x^{2}+7x-8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $7$ for $b$, and $-8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Square $7$.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-7±\sqrt{49+4\left(-8\right)}}{2\left(-1\right)}$$

Multiply $4$ times $-8$.

$$x=\frac{-7±\sqrt{49-32}}{2\left(-1\right)}$$

Add $49$ to $-32$.

$$x=\frac{-7±\sqrt{17}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-7±\sqrt{17}}{-2}$$

Now solve the equation $x=\frac{-7±\sqrt{17}}{-2}$ when $±$ is plus. Add $-7$ to $\sqrt{17}$.

$$x=\frac{\sqrt{17}-7}{-2}$$

Divide $-7+\sqrt{17}$ by $-2$.

$$x=\frac{7-\sqrt{17}}{2}$$

Now solve the equation $x=\frac{-7±\sqrt{17}}{-2}$ when $±$ is minus. Subtract $\sqrt{17}$ from $-7$.

$$x=\frac{-\sqrt{17}-7}{-2}$$

Divide $-7-\sqrt{17}$ by $-2$.

$$x=\frac{\sqrt{17}+7}{2}$$

The equation is now solved.

$$x=\frac{7-\sqrt{17}}{2}$$ $$x=\frac{\sqrt{17}+7}{2}$$

Steps for Completing the Square

Subtract $x^{2}$ from both sides.

$$x+1-x^{2}=-6x+9$$

Add $6x$ to both sides.

$$x+1-x^{2}+6x=9$$

Combine $x$ and $6x$ to get $7x$.

$$7x+1-x^{2}=9$$

Subtract $1$ from both sides.

$$7x-x^{2}=9-1$$

Subtract $1$ from $9$ to get $8$.

$$7x-x^{2}=8$$

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 $x^{2}+bx=c$.

$$-x^{2}+7x=8$$

Divide both sides by $-1$.

$$\frac{-x^{2}+7x}{-1}=\frac{8}{-1}$$

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

$$x^{2}+\frac{7}{-1}x=\frac{8}{-1}$$

Divide $7$ by $-1$.

$$x^{2}-7x=\frac{8}{-1}$$

Divide $8$ by $-1$.

$$x^{2}-7x=-8$$

Divide $-7$, the coefficient of the $x$ term, by $2$ to get $-\frac{7}{2}$. Then add the square of $-\frac{7}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-8+\left(-\frac{7}{2}\right)^{2}$$

Square $-\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-8+\frac{49}{4}$$

Add $-8$ to $\frac{49}{4}$.

$$x^{2}-7x+\frac{49}{4}=\frac{17}{4}$$

Factor $x^{2}-7x+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{17}}{2}$$ $$x-\frac{7}{2}=-\frac{\sqrt{17}}{2}$$

Add $\frac{7}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{17}+7}{2}$$ $$x=\frac{7-\sqrt{17}}{2}$$

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Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square