Solve \sqrt { 10 x + 6 } = x + 3 | Equatix

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Question

$$\sqrt { 10 x + 6 } = x + 3$$

Solve for x

$x=3$
$x=1$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{10x+6}\right)^{2}=\left(x+3\right)^{2}$$

Calculate $\sqrt{10x+6}$ to the power of $2$ and get $10x+6$.

$$10x+6=\left(x+3\right)^{2}$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+3\right)^{2}$.

$$10x+6=x^{2}+6x+9$$

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

$$10x+6-x^{2}=6x+9$$

Subtract $6x$ from both sides.

$$10x+6-x^{2}-6x=9$$

Combine $10x$ and $-6x$ to get $4x$.

$$4x+6-x^{2}=9$$

Subtract $9$ from both sides.

$$4x+6-x^{2}-9=0$$

Subtract $9$ from $6$ to get $-3$.

$$4x-3-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+4x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=-\left(-3\right)=3$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.

$$a=3$$ $$b=1$$

Rewrite $-x^{2}+4x-3$ as $\left(-x^{2}+3x\right)+\left(x-3\right)$.

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

Factor out $-x$ in $-x^{2}+3x$.

$$-x\left(x-3\right)+x-3$$

Factor out common term $x-3$ by using distributive property.

$$\left(x-3\right)\left(-x+1\right)$$

To find equation solutions, solve $x-3=0$ and $-x+1=0$.

$$x=3$$ $$x=1$$

Substitute $3$ for $x$ in the equation $\sqrt{10x+6}=x+3$.

$$\sqrt{10\times 3+6}=3+3$$

Simplify. The value $x=3$ satisfies the equation.

$$6=6$$

Substitute $1$ for $x$ in the equation $\sqrt{10x+6}=x+3$.

$$\sqrt{10\times 1+6}=1+3$$

Simplify. The value $x=1$ satisfies the equation.

$$4=4$$

List all solutions of $\sqrt{10x+6}=x+3$.