Solve 4x+13+4+4sqrt{4x+13}=-2x+3

$$4 x + 13 + 4 + 4 \sqrt { 4 x + 13 } = - 2 x + 3$$

Solve for x

$$x=-3$$

Solution Steps

Subtract 4x+13+4 from both sides of the equation.

$$4\sqrt{4x+13}=-2x+3-\left(4x+13+4\right)$$

Add 13 and 4 to get 17.

$$4\sqrt{4x+13}=-2x+3-\left(4x+17\right)$$

To find the opposite of 4x+17, find the opposite of each term.

$$4\sqrt{4x+13}=-2x+3-4x-17$$

Combine -2x and -4x to get -6x.

$$4\sqrt{4x+13}=-6x+3-17$$

Subtract 17 from 3 to get -14.

$$4\sqrt{4x+13}=-6x-14$$

Square both sides of the equation.

$$\left(4\sqrt{4x+13}\right)^{2}=\left(-6x-14\right)^{2}$$

Expand \left(4\sqrt{4x+13}\right)^{2}.

$$4^{2}\left(\sqrt{4x+13}\right)^{2}=\left(-6x-14\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16\left(\sqrt{4x+13}\right)^{2}=\left(-6x-14\right)^{2}$$

Calculate \sqrt{4x+13} to the power of 2 and get 4x+13.

$$16\left(4x+13\right)=\left(-6x-14\right)^{2}$$

Use the distributive property to multiply 16 by 4x+13.

$$64x+208=\left(-6x-14\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6x-14\right)^{2}.

$$64x+208=36x^{2}+168x+196$$

Subtract 36x^{2} from both sides.

$$64x+208-36x^{2}=168x+196$$

Subtract 168x from both sides.

$$64x+208-36x^{2}-168x=196$$

Combine 64x and -168x to get -104x.

$$-104x+208-36x^{2}=196$$

Subtract 196 from both sides.

$$-104x+208-36x^{2}-196=0$$

Subtract 196 from 208 to get 12.

$$-104x+12-36x^{2}=0$$

Divide both sides by 4.

$$-26x+3-9x^{2}=0$$

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

$$-9x^{2}-26x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-26 ab=-9\times 3=-27$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -27.

$$1,-27 3,-9$$

Calculate the sum for each pair.

$$1-27=-26 3-9=-6$$

The solution is the pair that gives sum -26.

$$a=1 b=-27$$

Rewrite -9x^{2}-26x+3 as \left(-9x^{2}+x\right)+\left(-27x+3\right).

$$\left(-9x^{2}+x\right)+\left(-27x+3\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(9x-1\right)-3\left(9x-1\right)$$

Factor out common term 9x-1 by using distributive property.

$$\left(9x-1\right)\left(-x-3\right)$$

To find equation solutions, solve 9x-1=0 and -x-3=0.

$$x=\frac{1}{9} x=-3$$

Substitute \frac{1}{9} for x in the equation 4x+13+4+4\sqrt{4x+13}=-2x+3.

$$4\times \frac{1}{9}+13+4+4\sqrt{4\times \frac{1}{9}+13}=-2\times \frac{1}{9}+3$$

Simplify. The value x=\frac{1}{9} does not satisfy the equation.

$$\frac{289}{9}=\frac{25}{9}$$

Substitute -3 for x in the equation 4x+13+4+4\sqrt{4x+13}=-2x+3.

$$4\left(-3\right)+13+4+4\sqrt{4\left(-3\right)+13}=-2\left(-3\right)+3$$

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

$$9=9$$

Equation 4\sqrt{4x+13}=-6x-14 has a unique solution.

$$x=-3$$