Solve x=17sqrt{x}-7

$$x = 17 \sqrt { x } - 7$$

Solve for x

$$x = \frac{51 \sqrt{29} + 275}{2} \approx 274.821702582$$

$$x=\frac{275-51\sqrt{29}}{2}\approx 0.178297418$$

Solution Steps

Subtract -7 from both sides of the equation.

$$x+7=17\sqrt{x}$$

Square both sides of the equation.

$$\left(x+7\right)^{2}=\left(17\sqrt{x}\right)^{2}$$

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

$$x^{2}+14x+49=\left(17\sqrt{x}\right)^{2}$$

Expand \left(17\sqrt{x}\right)^{2}.

$$x^{2}+14x+49=17^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 17 to the power of 2 and get 289.

$$x^{2}+14x+49=289\left(\sqrt{x}\right)^{2}$$

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

$$x^{2}+14x+49=289x$$

Subtract 289x from both sides.

$$x^{2}+14x+49-289x=0$$

Combine 14x and -289x to get -275x.

$$x^{2}-275x+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -275 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-275\right)±\sqrt{\left(-275\right)^{2}-4\times 49}}{2}$$

Square -275.

$$x=\frac{-\left(-275\right)±\sqrt{75625-4\times 49}}{2}$$

Multiply -4 times 49.

$$x=\frac{-\left(-275\right)±\sqrt{75625-196}}{2}$$

Add 75625 to -196.

$$x=\frac{-\left(-275\right)±\sqrt{75429}}{2}$$

Take the square root of 75429.

$$x=\frac{-\left(-275\right)±51\sqrt{29}}{2}$$

The opposite of -275 is 275.

$$x=\frac{275±51\sqrt{29}}{2}$$

Now solve the equation x=\frac{275±51\sqrt{29}}{2} when ± is plus. Add 275 to 51\sqrt{29}.

$$x=\frac{51\sqrt{29}+275}{2}$$

Now solve the equation x=\frac{275±51\sqrt{29}}{2} when ± is minus. Subtract 51\sqrt{29} from 275.

$$x=\frac{275-51\sqrt{29}}{2}$$

The equation is now solved.

$$x=\frac{51\sqrt{29}+275}{2} x=\frac{275-51\sqrt{29}}{2}$$

Substitute \frac{51\sqrt{29}+275}{2} for x in the equation x=17\sqrt{x}-7.

$$\frac{51\sqrt{29}+275}{2}=17\sqrt{\frac{51\sqrt{29}+275}{2}}-7$$

Simplify. The value x=\frac{51\sqrt{29}+275}{2} satisfies the equation.

$$\frac{51}{2}\times 29^{\frac{1}{2}}+\frac{275}{2}=\frac{275}{2}+\frac{51}{2}\times 29^{\frac{1}{2}}$$

Substitute \frac{275-51\sqrt{29}}{2} for x in the equation x=17\sqrt{x}-7.

$$\frac{275-51\sqrt{29}}{2}=17\sqrt{\frac{275-51\sqrt{29}}{2}}-7$$

Simplify. The value x=\frac{275-51\sqrt{29}}{2} satisfies the equation.

$$\frac{275}{2}-\frac{51}{2}\times 29^{\frac{1}{2}}=\frac{275}{2}-\frac{51}{2}\times 29^{\frac{1}{2}}$$

List all solutions of x+7=17\sqrt{x}.

$$x=\frac{51\sqrt{29}+275}{2} x=\frac{275-51\sqrt{29}}{2}$$