Solve (x)=sqrt{9x^2-28x+3}

$$( x ) = \sqrt { 9 x ^ { 2 } - 28 x + 3 }$$

Solve for x

$$x = \frac{\sqrt{43} + 7}{4} \approx 3.389359631$$

$$x=\frac{7-\sqrt{43}}{4}\approx 0.110640369$$

Solution Steps

Square both sides of the equation.

$$x^{2}=\left(\sqrt{9x^{2}-28x+3}\right)^{2}$$

Calculate \sqrt{9x^{2}-28x+3} to the power of 2 and get 9x^{2}-28x+3.

$$x^{2}=9x^{2}-28x+3$$

Subtract 9x^{2} from both sides.

$$x^{2}-9x^{2}=-28x+3$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}=-28x+3$$

Add 28x to both sides.

$$-8x^{2}+28x=3$$

Subtract 3 from both sides.

$$-8x^{2}+28x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 28 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\left(-8\right)\left(-3\right)}}{2\left(-8\right)}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\left(-8\right)\left(-3\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-28±\sqrt{784+32\left(-3\right)}}{2\left(-8\right)}$$

Multiply 32 times -3.

$$x=\frac{-28±\sqrt{784-96}}{2\left(-8\right)}$$

Add 784 to -96.

$$x=\frac{-28±\sqrt{688}}{2\left(-8\right)}$$

Take the square root of 688.

$$x=\frac{-28±4\sqrt{43}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-28±4\sqrt{43}}{-16}$$

Now solve the equation x=\frac{-28±4\sqrt{43}}{-16} when ± is plus. Add -28 to 4\sqrt{43}.

$$x=\frac{4\sqrt{43}-28}{-16}$$

Divide -28+4\sqrt{43} by -16.

$$x=\frac{7-\sqrt{43}}{4}$$

Now solve the equation x=\frac{-28±4\sqrt{43}}{-16} when ± is minus. Subtract 4\sqrt{43} from -28.

$$x=\frac{-4\sqrt{43}-28}{-16}$$

Divide -28-4\sqrt{43} by -16.

$$x=\frac{\sqrt{43}+7}{4}$$

The equation is now solved.

$$x=\frac{7-\sqrt{43}}{4} x=\frac{\sqrt{43}+7}{4}$$

Substitute \frac{7-\sqrt{43}}{4} for x in the equation x=\sqrt{9x^{2}-28x+3}.

$$\frac{7-\sqrt{43}}{4}=\sqrt{9\times \left(\frac{7-\sqrt{43}}{4}\right)^{2}-28\times \frac{7-\sqrt{43}}{4}+3}$$

Simplify. The value x=\frac{7-\sqrt{43}}{4} satisfies the equation.

$$\frac{7}{4}-\frac{1}{4}\times 43^{\frac{1}{2}}=\frac{7}{4}-\frac{1}{4}\times 43^{\frac{1}{2}}$$

Substitute \frac{\sqrt{43}+7}{4} for x in the equation x=\sqrt{9x^{2}-28x+3}.

$$\frac{\sqrt{43}+7}{4}=\sqrt{9\times \left(\frac{\sqrt{43}+7}{4}\right)^{2}-28\times \frac{\sqrt{43}+7}{4}+3}$$

Simplify. The value x=\frac{\sqrt{43}+7}{4} satisfies the equation.

$$\frac{1}{4}\times 43^{\frac{1}{2}}+\frac{7}{4}=\frac{7}{4}+\frac{1}{4}\times 43^{\frac{1}{2}}$$

List all solutions of x=\sqrt{9x^{2}-28x+3}.

$$x=\frac{7-\sqrt{43}}{4} x=\frac{\sqrt{43}+7}{4}$$