Solve 7x-3=8sqrt{x}

$$7x-3 = 8 \sqrt{ x }$$

Solve for x

$$x = \frac{8 \sqrt{37} + 53}{49} \approx 2.07473674$$

Solution Steps

Square both sides of the equation.

$$\left(7x-3\right)^{2}=\left(8\sqrt{x}\right)^{2}$$

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

$$49x^{2}-42x+9=\left(8\sqrt{x}\right)^{2}$$

Expand \left(8\sqrt{x}\right)^{2}.

$$49x^{2}-42x+9=8^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 8 to the power of 2 and get 64.

$$49x^{2}-42x+9=64\left(\sqrt{x}\right)^{2}$$

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

$$49x^{2}-42x+9=64x$$

Subtract 64x from both sides.

$$49x^{2}-42x+9-64x=0$$

Combine -42x and -64x to get -106x.

$$49x^{2}-106x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -106 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-106\right)±\sqrt{\left(-106\right)^{2}-4\times 49\times 9}}{2\times 49}$$

Square -106.

$$x=\frac{-\left(-106\right)±\sqrt{11236-4\times 49\times 9}}{2\times 49}$$

Multiply -4 times 49.

$$x=\frac{-\left(-106\right)±\sqrt{11236-196\times 9}}{2\times 49}$$

Multiply -196 times 9.

$$x=\frac{-\left(-106\right)±\sqrt{11236-1764}}{2\times 49}$$

Add 11236 to -1764.

$$x=\frac{-\left(-106\right)±\sqrt{9472}}{2\times 49}$$

Take the square root of 9472.

$$x=\frac{-\left(-106\right)±16\sqrt{37}}{2\times 49}$$

The opposite of -106 is 106.

$$x=\frac{106±16\sqrt{37}}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{106±16\sqrt{37}}{98}$$

Now solve the equation x=\frac{106±16\sqrt{37}}{98} when ± is plus. Add 106 to 16\sqrt{37}.

$$x=\frac{16\sqrt{37}+106}{98}$$

Divide 106+16\sqrt{37} by 98.

$$x=\frac{8\sqrt{37}+53}{49}$$

Now solve the equation x=\frac{106±16\sqrt{37}}{98} when ± is minus. Subtract 16\sqrt{37} from 106.

$$x=\frac{106-16\sqrt{37}}{98}$$

Divide 106-16\sqrt{37} by 98.

$$x=\frac{53-8\sqrt{37}}{49}$$

The equation is now solved.

$$x=\frac{8\sqrt{37}+53}{49} x=\frac{53-8\sqrt{37}}{49}$$

Substitute \frac{8\sqrt{37}+53}{49} for x in the equation 7x-3=8\sqrt{x}.

$$7\times \frac{8\sqrt{37}+53}{49}-3=8\sqrt{\frac{8\sqrt{37}+53}{49}}$$

Simplify. The value x=\frac{8\sqrt{37}+53}{49} satisfies the equation.

$$\frac{8}{7}\times 37^{\frac{1}{2}}+\frac{32}{7}=\frac{8}{7}\times 37^{\frac{1}{2}}+\frac{32}{7}$$

Substitute \frac{53-8\sqrt{37}}{49} for x in the equation 7x-3=8\sqrt{x}.

$$7\times \frac{53-8\sqrt{37}}{49}-3=8\sqrt{\frac{53-8\sqrt{37}}{49}}$$

Simplify. The value x=\frac{53-8\sqrt{37}}{49} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{32}{7}-\frac{8}{7}\times 37^{\frac{1}{2}}=\frac{8}{7}\times 37^{\frac{1}{2}}-\frac{32}{7}$$

Equation 7x-3=8\sqrt{x} has a unique solution.

$$x=\frac{8\sqrt{37}+53}{49}$$