Solve *3) ( iv ) (x - 3)(2x + 1) = x(x + 5) | 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

$$( x - 3 ) ( 2 x + 1 ) = x ( x + 5 )$$

Solve for x

$x=2\sqrt{7}+5\approx 10.291502622$
$x=5-2\sqrt{7}\approx -0.291502622$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $x-3$ by $2x+1$ and combine like terms.

$$2x^{2}-5x-3=x\left(x+5\right)$$

Use the distributive property to multiply $x$ by $x+5$.

$$2x^{2}-5x-3=x^{2}+5x$$

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

$$2x^{2}-5x-3-x^{2}=5x$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}-5x-3=5x$$

Subtract $5x$ from both sides.

$$x^{2}-5x-3-5x=0$$

Combine $-5x$ and $-5x$ to get $-10x$.

$$x^{2}-10x-3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-10$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-3\right)}}{2}$$

Square $-10$.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)}}{2}$$

Multiply $-4$ times $-3$.

$$x=\frac{-\left(-10\right)±\sqrt{100+12}}{2}$$

Add $100$ to $12$.

$$x=\frac{-\left(-10\right)±\sqrt{112}}{2}$$

Take the square root of $112$.

$$x=\frac{-\left(-10\right)±4\sqrt{7}}{2}$$

The opposite of $-10$ is $10$.

$$x=\frac{10±4\sqrt{7}}{2}$$

Now solve the equation $x=\frac{10±4\sqrt{7}}{2}$ when $±$ is plus. Add $10$ to $4\sqrt{7}$.

$$x=\frac{4\sqrt{7}+10}{2}$$

Divide $10+4\sqrt{7}$ by $2$.

$$x=2\sqrt{7}+5$$

Now solve the equation $x=\frac{10±4\sqrt{7}}{2}$ when $±$ is minus. Subtract $4\sqrt{7}$ from $10$.

$$x=\frac{10-4\sqrt{7}}{2}$$

Divide $10-4\sqrt{7}$ by $2$.

$$x=5-2\sqrt{7}$$

The equation is now solved.

$$x=2\sqrt{7}+5$$ $$x=5-2\sqrt{7}$$

Steps for Completing the Square

Use the distributive property to multiply $x-3$ by $2x+1$ and combine like terms.

$$2x^{2}-5x-3=x\left(x+5\right)$$

Use the distributive property to multiply $x$ by $x+5$.

$$2x^{2}-5x-3=x^{2}+5x$$

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

$$2x^{2}-5x-3-x^{2}=5x$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}-5x-3=5x$$

Subtract $5x$ from both sides.

$$x^{2}-5x-3-5x=0$$

Combine $-5x$ and $-5x$ to get $-10x$.

$$x^{2}-10x-3=0$$

Add $3$ to both sides. Anything plus zero gives itself.

$$x^{2}-10x=3$$

Divide $-10$, the coefficient of the $x$ term, by $2$ to get $-5$. Then add the square of $-5$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=3+\left(-5\right)^{2}$$

Square $-5$.

$$x^{2}-10x+25=3+25$$

Add $3$ to $25$.

$$x^{2}-10x+25=28$$

Factor $x^{2}-10x+25$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-5\right)^{2}=28$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{28}$$

Simplify.

$$x-5=2\sqrt{7}$$ $$x-5=-2\sqrt{7}$$

Add $5$ to both sides of the equation.

$$x=2\sqrt{7}+5$$ $$x=5-2\sqrt{7}$$