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

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

Solve for x

$$x = \frac{39 - 5 \sqrt{41}}{2} \approx 3.492189406$$

Solution Steps

Square both sides of the equation.

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

Expand \left(5\sqrt{x-3}\right)^{2}.

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

Calculate 5 to the power of 2 and get 25.

$$25\left(\sqrt{x-3}\right)^{2}=\left(7-x\right)^{2}$$

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

$$25\left(x-3\right)=\left(7-x\right)^{2}$$

Use the distributive property to multiply 25 by x-3.

$$25x-75=\left(7-x\right)^{2}$$

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

$$25x-75=49-14x+x^{2}$$

Subtract 49 from both sides.

$$25x-75-49=-14x+x^{2}$$

Subtract 49 from -75 to get -124.

$$25x-124=-14x+x^{2}$$

Add 14x to both sides.

$$25x-124+14x=x^{2}$$

Combine 25x and 14x to get 39x.

$$39x-124=x^{2}$$

Subtract x^{2} from both sides.

$$39x-124-x^{2}=0$$

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.

$$-x^{2}+39x-124=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 39 for b, and -124 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-39±\sqrt{39^{2}-4\left(-1\right)\left(-124\right)}}{2\left(-1\right)}$$

Square 39.

$$x=\frac{-39±\sqrt{1521-4\left(-1\right)\left(-124\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-39±\sqrt{1521+4\left(-124\right)}}{2\left(-1\right)}$$

Multiply 4 times -124.

$$x=\frac{-39±\sqrt{1521-496}}{2\left(-1\right)}$$

Add 1521 to -496.

$$x=\frac{-39±\sqrt{1025}}{2\left(-1\right)}$$

Take the square root of 1025.

$$x=\frac{-39±5\sqrt{41}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-39±5\sqrt{41}}{-2}$$

Now solve the equation x=\frac{-39±5\sqrt{41}}{-2} when ± is plus. Add -39 to 5\sqrt{41}.

$$x=\frac{5\sqrt{41}-39}{-2}$$

Divide -39+5\sqrt{41} by -2.

$$x=\frac{39-5\sqrt{41}}{2}$$

Now solve the equation x=\frac{-39±5\sqrt{41}}{-2} when ± is minus. Subtract 5\sqrt{41} from -39.

$$x=\frac{-5\sqrt{41}-39}{-2}$$

Divide -39-5\sqrt{41} by -2.

$$x=\frac{5\sqrt{41}+39}{2}$$

The equation is now solved.

$$x=\frac{39-5\sqrt{41}}{2} x=\frac{5\sqrt{41}+39}{2}$$

Substitute \frac{39-5\sqrt{41}}{2} for x in the equation 5\sqrt{x-3}=7-x.

$$5\sqrt{\frac{39-5\sqrt{41}}{2}-3}=7-\frac{39-5\sqrt{41}}{2}$$

Simplify. The value x=\frac{39-5\sqrt{41}}{2} satisfies the equation.

$$-\frac{25}{2}+\frac{5}{2}\times 41^{\frac{1}{2}}=-\frac{25}{2}+\frac{5}{2}\times 41^{\frac{1}{2}}$$

Substitute \frac{5\sqrt{41}+39}{2} for x in the equation 5\sqrt{x-3}=7-x.

$$5\sqrt{\frac{5\sqrt{41}+39}{2}-3}=7-\frac{5\sqrt{41}+39}{2}$$

Simplify. The value x=\frac{5\sqrt{41}+39}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{25}{2}+\frac{5}{2}\times 41^{\frac{1}{2}}=-\frac{25}{2}-\frac{5}{2}\times 41^{\frac{1}{2}}$$

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

$$x=\frac{39-5\sqrt{41}}{2}$$