Solve sqrt{7-2x}-sqrt{5+x}=sqrt{4+3x}

$$\sqrt{ 7-2x } - \sqrt{ 5+x } = \sqrt{ 4+3x }$$

Solve for x

$$x=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{7-2x}-\sqrt{5+x}\right)^{2}=\left(\sqrt{4+3x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7-2x}-\sqrt{5+x}\right)^{2}.

$$\left(\sqrt{7-2x}\right)^{2}-2\sqrt{7-2x}\sqrt{5+x}+\left(\sqrt{5+x}\right)^{2}=\left(\sqrt{4+3x}\right)^{2}$$

Calculate \sqrt{7-2x} to the power of 2 and get 7-2x.

$$7-2x-2\sqrt{7-2x}\sqrt{5+x}+\left(\sqrt{5+x}\right)^{2}=\left(\sqrt{4+3x}\right)^{2}$$

Calculate \sqrt{5+x} to the power of 2 and get 5+x.

$$7-2x-2\sqrt{7-2x}\sqrt{5+x}+5+x=\left(\sqrt{4+3x}\right)^{2}$$

Add 7 and 5 to get 12.

$$12-2x-2\sqrt{7-2x}\sqrt{5+x}+x=\left(\sqrt{4+3x}\right)^{2}$$

Combine -2x and x to get -x.

$$12-x-2\sqrt{7-2x}\sqrt{5+x}=\left(\sqrt{4+3x}\right)^{2}$$

Calculate \sqrt{4+3x} to the power of 2 and get 4+3x.

$$12-x-2\sqrt{7-2x}\sqrt{5+x}=4+3x$$

Subtract 12-x from both sides of the equation.

$$-2\sqrt{7-2x}\sqrt{5+x}=4+3x-\left(12-x\right)$$

To find the opposite of 12-x, find the opposite of each term.

$$-2\sqrt{7-2x}\sqrt{5+x}=4+3x-12+x$$

Subtract 12 from 4 to get -8.

$$-2\sqrt{7-2x}\sqrt{5+x}=-8+3x+x$$

Combine 3x and x to get 4x.

$$-2\sqrt{7-2x}\sqrt{5+x}=-8+4x$$

Square both sides of the equation.

$$\left(-2\sqrt{7-2x}\sqrt{5+x}\right)^{2}=\left(-8+4x\right)^{2}$$

Expand \left(-2\sqrt{7-2x}\sqrt{5+x}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{7-2x}\right)^{2}\left(\sqrt{5+x}\right)^{2}=\left(-8+4x\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4\left(\sqrt{7-2x}\right)^{2}\left(\sqrt{5+x}\right)^{2}=\left(-8+4x\right)^{2}$$

Calculate \sqrt{7-2x} to the power of 2 and get 7-2x.

$$4\left(7-2x\right)\left(\sqrt{5+x}\right)^{2}=\left(-8+4x\right)^{2}$$

Calculate \sqrt{5+x} to the power of 2 and get 5+x.

$$4\left(7-2x\right)\left(5+x\right)=\left(-8+4x\right)^{2}$$

Use the distributive property to multiply 4 by 7-2x.

$$\left(28-8x\right)\left(5+x\right)=\left(-8+4x\right)^{2}$$

Apply the distributive property by multiplying each term of 28-8x by each term of 5+x.

$$140+28x-40x-8x^{2}=\left(-8+4x\right)^{2}$$

Combine 28x and -40x to get -12x.

$$140-12x-8x^{2}=\left(-8+4x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-8+4x\right)^{2}.

$$140-12x-8x^{2}=64-64x+16x^{2}$$

Subtract 64 from both sides.

$$140-12x-8x^{2}-64=-64x+16x^{2}$$

Subtract 64 from 140 to get 76.

$$76-12x-8x^{2}=-64x+16x^{2}$$

Add 64x to both sides.

$$76-12x-8x^{2}+64x=16x^{2}$$

Combine -12x and 64x to get 52x.

$$76+52x-8x^{2}=16x^{2}$$

Subtract 16x^{2} from both sides.

$$76+52x-8x^{2}-16x^{2}=0$$

Combine -8x^{2} and -16x^{2} to get -24x^{2}.

$$76+52x-24x^{2}=0$$

Divide both sides by 4.

$$19+13x-6x^{2}=0$$

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

$$-6x^{2}+13x+19=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -6x^{2}+ax+bx+19. To find a and b, set up a system to be solved.

$$a+b=13 ab=-6\times 19=-114$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -114.

$$-1,114 -2,57 -3,38 -6,19$$

Calculate the sum for each pair.

$$-1+114=113 -2+57=55 -3+38=35 -6+19=13$$

The solution is the pair that gives sum 13.

$$a=19 b=-6$$

Rewrite -6x^{2}+13x+19 as \left(-6x^{2}+19x\right)+\left(-6x+19\right).

$$\left(-6x^{2}+19x\right)+\left(-6x+19\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(6x-19\right)-\left(6x-19\right)$$

Factor out common term 6x-19 by using distributive property.

$$\left(6x-19\right)\left(-x-1\right)$$

To find equation solutions, solve 6x-19=0 and -x-1=0.

$$x=\frac{19}{6} x=-1$$

Substitute \frac{19}{6} for x in the equation \sqrt{7-2x}-\sqrt{5+x}=\sqrt{4+3x}.

$$\sqrt{7-2\times \frac{19}{6}}-\sqrt{5+\frac{19}{6}}=\sqrt{4+3\times \frac{19}{6}}$$

Simplify. The value x=\frac{19}{6} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{5}{6}\times 6^{\frac{1}{2}}=\frac{3}{2}\times 6^{\frac{1}{2}}$$

Substitute -1 for x in the equation \sqrt{7-2x}-\sqrt{5+x}=\sqrt{4+3x}.

$$\sqrt{7-2\left(-1\right)}-\sqrt{5-1}=\sqrt{4+3\left(-1\right)}$$

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

$$1=1$$

Equation -\sqrt{x+5}+\sqrt{7-2x}=\sqrt{3x+4} has a unique solution.

$$x=-1$$