Solve عمال 2 - x ^ 2 = 7x; i) | 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

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

Solve for x

$x=\frac{\sqrt{57}-7}{2}\approx 0.274917218$
$x=\frac{-\sqrt{57}-7}{2}\approx -7.274917218$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $7x$ from both sides.

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

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.

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

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

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

Square $-7$.

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

Multiply $-4$ times $-1$.

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

Multiply $4$ times $2$.

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

Add $49$ to $8$.

$$x=\frac{-\left(-7\right)±\sqrt{57}}{2\left(-1\right)}$$

The opposite of $-7$ is $7$.

$$x=\frac{7±\sqrt{57}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{7±\sqrt{57}}{-2}$$

Now solve the equation $x=\frac{7±\sqrt{57}}{-2}$ when $±$ is plus. Add $7$ to $\sqrt{57}$.

$$x=\frac{\sqrt{57}+7}{-2}$$

Divide $7+\sqrt{57}$ by $-2$.

$$x=\frac{-\sqrt{57}-7}{2}$$

Now solve the equation $x=\frac{7±\sqrt{57}}{-2}$ when $±$ is minus. Subtract $\sqrt{57}$ from $7$.

$$x=\frac{7-\sqrt{57}}{-2}$$

Divide $7-\sqrt{57}$ by $-2$.

$$x=\frac{\sqrt{57}-7}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{57}-7}{2}$$ $$x=\frac{\sqrt{57}-7}{2}$$

Steps for Completing the Square

Subtract $7x$ from both sides.

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

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

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

Divide both sides by $-1$.

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

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

$$x^{2}+\left(-\frac{7}{-1}\right)x=-\frac{2}{-1}$$

Divide $-7$ by $-1$.

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

Divide $-2$ by $-1$.

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

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.

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

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

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

Add $2$ to $\frac{49}{4}$.

$$x^{2}+7x+\frac{49}{4}=\frac{57}{4}$$

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

$$\left(x+\frac{7}{2}\right)^{2}=\frac{57}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{57}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{\sqrt{57}}{2}$$ $$x+\frac{7}{2}=-\frac{\sqrt{57}}{2}$$

Subtract $\frac{7}{2}$ from both sides of the equation.

$$x=\frac{\sqrt{57}-7}{2}$$ $$x=\frac{-\sqrt{57}-7}{2}$$