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

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

Solve for x

$$x=7$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7-x}+3\right)^{2}.

$$\left(\sqrt{7-x}\right)^{2}+6\sqrt{7-x}+9=\left(\sqrt{2x-5}\right)^{2}$$

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

$$7-x+6\sqrt{7-x}+9=\left(\sqrt{2x-5}\right)^{2}$$

Add 7 and 9 to get 16.

$$16-x+6\sqrt{7-x}=\left(\sqrt{2x-5}\right)^{2}$$

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

$$16-x+6\sqrt{7-x}=2x-5$$

Subtract 16-x from both sides of the equation.

$$6\sqrt{7-x}=2x-5-\left(16-x\right)$$

To find the opposite of 16-x, find the opposite of each term.

$$6\sqrt{7-x}=2x-5-16+x$$

Subtract 16 from -5 to get -21.

$$6\sqrt{7-x}=2x-21+x$$

Combine 2x and x to get 3x.

$$6\sqrt{7-x}=3x-21$$

Square both sides of the equation.

$$\left(6\sqrt{7-x}\right)^{2}=\left(3x-21\right)^{2}$$

Expand \left(6\sqrt{7-x}\right)^{2}.

$$6^{2}\left(\sqrt{7-x}\right)^{2}=\left(3x-21\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36\left(\sqrt{7-x}\right)^{2}=\left(3x-21\right)^{2}$$

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

$$36\left(7-x\right)=\left(3x-21\right)^{2}$$

Use the distributive property to multiply 36 by 7-x.

$$252-36x=\left(3x-21\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-21\right)^{2}.

$$252-36x=9x^{2}-126x+441$$

Subtract 9x^{2} from both sides.

$$252-36x-9x^{2}=-126x+441$$

Add 126x to both sides.

$$252-36x-9x^{2}+126x=441$$

Combine -36x and 126x to get 90x.

$$252+90x-9x^{2}=441$$

Subtract 441 from both sides.

$$252+90x-9x^{2}-441=0$$

Subtract 441 from 252 to get -189.

$$-189+90x-9x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-9x^{2}+90x-189=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 90 for b, and -189 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-90±\sqrt{90^{2}-4\left(-9\right)\left(-189\right)}}{2\left(-9\right)}$$

Square 90.

$$x=\frac{-90±\sqrt{8100-4\left(-9\right)\left(-189\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-90±\sqrt{8100+36\left(-189\right)}}{2\left(-9\right)}$$

Multiply 36 times -189.

$$x=\frac{-90±\sqrt{8100-6804}}{2\left(-9\right)}$$

Add 8100 to -6804.

$$x=\frac{-90±\sqrt{1296}}{2\left(-9\right)}$$

Take the square root of 1296.

$$x=\frac{-90±36}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-90±36}{-18}$$

Now solve the equation x=\frac{-90±36}{-18} when ± is plus. Add -90 to 36.

$$x=-\frac{54}{-18}$$

Divide -54 by -18.

$$x=3$$

Now solve the equation x=\frac{-90±36}{-18} when ± is minus. Subtract 36 from -90.

$$x=-\frac{126}{-18}$$

Divide -126 by -18.

$$x=7$$

The equation is now solved.

$$x=3 x=7$$

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

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

Simplify. The value x=3 does not satisfy the equation.

$$5=1$$

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

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

Simplify. The value x=7 satisfies the equation.

$$3=3$$

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

$$x=7$$