Solve sqrt{9+14-x^2}=2*x-35/7

$$\sqrt { 9 + 14 - x ^ { 2 } } = \frac { 2 \cdot x - 35 } { 7 }$$

Solve for x (complex solution)

$$x=\frac{70+7\sqrt{6}i}{53}\approx 1.320754717+0.323517513i$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{9+14-x^{2}}\right)^{2}=\left(\frac{2x-35}{7}\right)^{2}$$

Add 9 and 14 to get 23.

$$\left(\sqrt{23-x^{2}}\right)^{2}=\left(\frac{2x-35}{7}\right)^{2}$$

Calculate \sqrt{23-x^{2}} to the power of 2 and get 23-x^{2}.

$$23-x^{2}=\left(\frac{2x-35}{7}\right)^{2}$$

To raise \frac{2x-35}{7} to a power, raise both numerator and denominator to the power and then divide.

$$23-x^{2}=\frac{\left(2x-35\right)^{2}}{7^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-35\right)^{2}.

$$23-x^{2}=\frac{4x^{2}-140x+1225}{7^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$23-x^{2}=\frac{4x^{2}-140x+1225}{49}$$

Divide each term of 4x^{2}-140x+1225 by 49 to get \frac{4}{49}x^{2}-\frac{20}{7}x+25.

$$23-x^{2}=\frac{4}{49}x^{2}-\frac{20}{7}x+25$$

Subtract \frac{4}{49}x^{2} from both sides.

$$23-x^{2}-\frac{4}{49}x^{2}=-\frac{20}{7}x+25$$

Combine -x^{2} and -\frac{4}{49}x^{2} to get -\frac{53}{49}x^{2}.

$$23-\frac{53}{49}x^{2}=-\frac{20}{7}x+25$$

Add \frac{20}{7}x to both sides.

$$23-\frac{53}{49}x^{2}+\frac{20}{7}x=25$$

Subtract 25 from both sides.

$$23-\frac{53}{49}x^{2}+\frac{20}{7}x-25=0$$

Subtract 25 from 23 to get -2.

$$-2-\frac{53}{49}x^{2}+\frac{20}{7}x=0$$

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.

$$-\frac{53}{49}x^{2}+\frac{20}{7}x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{53}{49} for a, \frac{20}{7} for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{20}{7}±\sqrt{\left(\frac{20}{7}\right)^{2}-4\left(-\frac{53}{49}\right)\left(-2\right)}}{2\left(-\frac{53}{49}\right)}$$

Square \frac{20}{7} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{20}{7}±\sqrt{\frac{400}{49}-4\left(-\frac{53}{49}\right)\left(-2\right)}}{2\left(-\frac{53}{49}\right)}$$

Multiply -4 times -\frac{53}{49}.

$$x=\frac{-\frac{20}{7}±\sqrt{\frac{400}{49}+\frac{212}{49}\left(-2\right)}}{2\left(-\frac{53}{49}\right)}$$

Multiply \frac{212}{49} times -2.

$$x=\frac{-\frac{20}{7}±\sqrt{\frac{400-424}{49}}}{2\left(-\frac{53}{49}\right)}$$

Add \frac{400}{49} to -\frac{424}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{20}{7}±\sqrt{-\frac{24}{49}}}{2\left(-\frac{53}{49}\right)}$$

Take the square root of -\frac{24}{49}.

$$x=\frac{-\frac{20}{7}±\frac{2\sqrt{6}i}{7}}{2\left(-\frac{53}{49}\right)}$$

Multiply 2 times -\frac{53}{49}.

$$x=\frac{-\frac{20}{7}±\frac{2\sqrt{6}i}{7}}{-\frac{106}{49}}$$

Now solve the equation x=\frac{-\frac{20}{7}±\frac{2\sqrt{6}i}{7}}{-\frac{106}{49}} when ± is plus. Add -\frac{20}{7} to \frac{2i\sqrt{6}}{7}.

$$x=\frac{-20+2\sqrt{6}i}{-\frac{106}{49}\times 7}$$

Divide \frac{-20+2i\sqrt{6}}{7} by -\frac{106}{49} by multiplying \frac{-20+2i\sqrt{6}}{7} by the reciprocal of -\frac{106}{49}.

$$x=\frac{-7\sqrt{6}i+70}{53}$$

Now solve the equation x=\frac{-\frac{20}{7}±\frac{2\sqrt{6}i}{7}}{-\frac{106}{49}} when ± is minus. Subtract \frac{2i\sqrt{6}}{7} from -\frac{20}{7}.

$$x=\frac{-2\sqrt{6}i-20}{-\frac{106}{49}\times 7}$$

Divide \frac{-20-2i\sqrt{6}}{7} by -\frac{106}{49} by multiplying \frac{-20-2i\sqrt{6}}{7} by the reciprocal of -\frac{106}{49}.

$$x=\frac{70+7\sqrt{6}i}{53}$$

The equation is now solved.

$$x=\frac{-7\sqrt{6}i+70}{53} x=\frac{70+7\sqrt{6}i}{53}$$

Substitute \frac{-7\sqrt{6}i+70}{53} for x in the equation \sqrt{9+14-x^{2}}=\frac{2x-35}{7}.

$$\sqrt{9+14-\left(\frac{-7\sqrt{6}i+70}{53}\right)^{2}}=\frac{2\times \frac{-7\sqrt{6}i+70}{53}-35}{7}$$

Simplify. The value x=\frac{-7\sqrt{6}i+70}{53} does not satisfy the equation.

$$\frac{245}{53}+\frac{2}{53}i\times 6^{\frac{1}{2}}=-\frac{2}{53}i\times 6^{\frac{1}{2}}-\frac{245}{53}$$

Substitute \frac{70+7\sqrt{6}i}{53} for x in the equation \sqrt{9+14-x^{2}}=\frac{2x-35}{7}.

$$\sqrt{9+14-\left(\frac{70+7\sqrt{6}i}{53}\right)^{2}}=\frac{2\times \frac{70+7\sqrt{6}i}{53}-35}{7}$$

Simplify. The value x=\frac{70+7\sqrt{6}i}{53} satisfies the equation.

$$-\frac{245}{53}+\frac{2}{53}i\times 6^{\frac{1}{2}}=-\frac{245}{53}+\frac{2}{53}i\times 6^{\frac{1}{2}}$$

Equation \sqrt{23-x^{2}}=\frac{2x-35}{7} has a unique solution.

$$x=\frac{70+7\sqrt{6}i}{53}$$