Solve sqrt{25-x^2}-sqrt{15+x^2}=4

$$\sqrt { 25 - x ^ { 2 } } - \sqrt { 15 + x ^ { 2 } } = 4$$

Solve for x (complex solution)

$$x=-\sqrt{11}i\approx -0-3.31662479i$$

$$x=\sqrt{11}i\approx 3.31662479i$$

Solution Steps

Subtract -\sqrt{15+x^{2}} from both sides of the equation.

$$\sqrt{25-x^{2}}=4+\sqrt{15+x^{2}}$$

Square both sides of the equation.

$$\left(\sqrt{25-x^{2}}\right)^{2}=\left(4+\sqrt{15+x^{2}}\right)^{2}$$

Calculate \sqrt{25-x^{2}} to the power of 2 and get 25-x^{2}.

$$25-x^{2}=\left(4+\sqrt{15+x^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+\sqrt{15+x^{2}}\right)^{2}.

$$25-x^{2}=16+8\sqrt{15+x^{2}}+\left(\sqrt{15+x^{2}}\right)^{2}$$

Calculate \sqrt{15+x^{2}} to the power of 2 and get 15+x^{2}.

$$25-x^{2}=16+8\sqrt{15+x^{2}}+15+x^{2}$$

Add 16 and 15 to get 31.

$$25-x^{2}=31+8\sqrt{15+x^{2}}+x^{2}$$

Subtract 31+x^{2} from both sides of the equation.

$$25-x^{2}-\left(31+x^{2}\right)=8\sqrt{15+x^{2}}$$

To find the opposite of 31+x^{2}, find the opposite of each term.

$$25-x^{2}-31-x^{2}=8\sqrt{15+x^{2}}$$

Subtract 31 from 25 to get -6.

$$-6-x^{2}-x^{2}=8\sqrt{15+x^{2}}$$

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

$$-6-2x^{2}=8\sqrt{15+x^{2}}$$

Square both sides of the equation.

$$\left(-6-2x^{2}\right)^{2}=\left(8\sqrt{15+x^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-6-2x^{2}\right)^{2}.

$$36+24x^{2}+4\left(x^{2}\right)^{2}=\left(8\sqrt{15+x^{2}}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$36+24x^{2}+4x^{4}=\left(8\sqrt{15+x^{2}}\right)^{2}$$

Expand \left(8\sqrt{15+x^{2}}\right)^{2}.

$$36+24x^{2}+4x^{4}=8^{2}\left(\sqrt{15+x^{2}}\right)^{2}$$

Calculate 8 to the power of 2 and get 64.

$$36+24x^{2}+4x^{4}=64\left(\sqrt{15+x^{2}}\right)^{2}$$

Calculate \sqrt{15+x^{2}} to the power of 2 and get 15+x^{2}.

$$36+24x^{2}+4x^{4}=64\left(15+x^{2}\right)$$

Use the distributive property to multiply 64 by 15+x^{2}.

$$36+24x^{2}+4x^{4}=960+64x^{2}$$

Subtract 960 from both sides.

$$36+24x^{2}+4x^{4}-960=64x^{2}$$

Subtract 960 from 36 to get -924.

$$-924+24x^{2}+4x^{4}=64x^{2}$$

Subtract 64x^{2} from both sides.

$$-924+24x^{2}+4x^{4}-64x^{2}=0$$

Combine 24x^{2} and -64x^{2} to get -40x^{2}.

$$-924-40x^{2}+4x^{4}=0$$

Substitute t for x^{2}.

$$4t^{2}-40t-924=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, -40 for b, and -924 for c in the quadratic formula.

$$t=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 4\left(-924\right)}}{2\times 4}$$

Do the calculations.

$$t=\frac{40±128}{8}$$

Solve the equation t=\frac{40±128}{8} when ± is plus and when ± is minus.

$$t=21 t=-11$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=-\sqrt{21} x=\sqrt{21} x=-\sqrt{11}i x=\sqrt{11}i$$

Substitute -\sqrt{21} for x in the equation \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(-\sqrt{21}\right)^{2}}-\sqrt{15+\left(-\sqrt{21}\right)^{2}}=4$$

Simplify. The value x=-\sqrt{21} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-4=4$$

Substitute \sqrt{21} for x in the equation \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(\sqrt{21}\right)^{2}}-\sqrt{15+\left(\sqrt{21}\right)^{2}}=4$$

Simplify. The value x=\sqrt{21} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-4=4$$

Substitute -\sqrt{11}i for x in the equation \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(-\sqrt{11}i\right)^{2}}-\sqrt{15+\left(-\sqrt{11}i\right)^{2}}=4$$

Simplify. The value x=-\sqrt{11}i satisfies the equation.

$$4=4$$

Substitute \sqrt{11}i for x in the equation \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(\sqrt{11}i\right)^{2}}-\sqrt{15+\left(\sqrt{11}i\right)^{2}}=4$$

Simplify. The value x=\sqrt{11}i satisfies the equation.

$$4=4$$

List all solutions of \sqrt{25-x^{2}}=\sqrt{x^{2}+15}+4.

$$x=-\sqrt{11}i x=\sqrt{11}i$$