Solve អង្គារ ពុធ សុ 2 * (x + 1) ^ 2 + 21 = 3 | 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

$$8 ( x + 1 ) ^ { 2 } + 21 = 3$$

Solve for x (complex solution)

$x=-1-\frac{3}{2}i=-1-1.5i$
$x=-1+\frac{3}{2}i=-1+1.5i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+1\right)^{2}$.

$$8\left(x^{2}+2x+1\right)+21=3$$

Use the distributive property to multiply $8$ by $x^{2}+2x+1$.

$$8x^{2}+16x+8+21=3$$

Add $8$ and $21$ to get $29$.

$$8x^{2}+16x+29=3$$

Subtract $3$ from both sides.

$$8x^{2}+16x+29-3=0$$

Subtract $3$ from $29$ to get $26$.

$$8x^{2}+16x+26=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $8$ for $a$, $16$ for $b$, and $26$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-16±\sqrt{16^{2}-4\times 8\times 26}}{2\times 8}$$

Square $16$.

$$x=\frac{-16±\sqrt{256-4\times 8\times 26}}{2\times 8}$$

Multiply $-4$ times $8$.

$$x=\frac{-16±\sqrt{256-32\times 26}}{2\times 8}$$

Multiply $-32$ times $26$.

$$x=\frac{-16±\sqrt{256-832}}{2\times 8}$$

Add $256$ to $-832$.

$$x=\frac{-16±\sqrt{-576}}{2\times 8}$$

Take the square root of $-576$.

$$x=\frac{-16±24i}{2\times 8}$$

Multiply $2$ times $8$.

$$x=\frac{-16±24i}{16}$$

Now solve the equation $x=\frac{-16±24i}{16}$ when $±$ is plus. Add $-16$ to $24i$.

$$x=\frac{-16+24i}{16}$$

Divide $-16+24i$ by $16$.

$$x=-1+\frac{3}{2}i$$

Now solve the equation $x=\frac{-16±24i}{16}$ when $±$ is minus. Subtract $24i$ from $-16$.

$$x=\frac{-16-24i}{16}$$

Divide $-16-24i$ by $16$.

$$x=-1-\frac{3}{2}i$$

The equation is now solved.

$$x=-1+\frac{3}{2}i$$ $$x=-1-\frac{3}{2}i$$

Steps for Completing the Square

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+1\right)^{2}$.

$$8\left(x^{2}+2x+1\right)+21=3$$

Use the distributive property to multiply $8$ by $x^{2}+2x+1$.

$$8x^{2}+16x+8+21=3$$

Add $8$ and $21$ to get $29$.

$$8x^{2}+16x+29=3$$

Subtract $29$ from both sides.

$$8x^{2}+16x=3-29$$

Subtract $29$ from $3$ to get $-26$.

$$8x^{2}+16x=-26$$

Divide both sides by $8$.

$$\frac{8x^{2}+16x}{8}=-\frac{26}{8}$$

Dividing by $8$ undoes the multiplication by $8$.

$$x^{2}+\frac{16}{8}x=-\frac{26}{8}$$

Divide $16$ by $8$.

$$x^{2}+2x=-\frac{26}{8}$$

Reduce the fraction $\frac{-26}{8}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}+2x=-\frac{13}{4}$$

Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=-\frac{13}{4}+1^{2}$$

Square $1$.

$$x^{2}+2x+1=-\frac{13}{4}+1$$

Add $-\frac{13}{4}$ to $1$.

$$x^{2}+2x+1=-\frac{9}{4}$$

Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+1\right)^{2}=-\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{-\frac{9}{4}}$$

Simplify.

$$x+1=\frac{3}{2}i$$ $$x+1=-\frac{3}{2}i$$

Subtract $1$ from both sides of the equation.

$$x=-1+\frac{3}{2}i$$ $$x=-1-\frac{3}{2}i$$