Solve 2x^2-9={left(x+sqrt{25-x^2}right)}^2-2

$$2 { x }^{ 2 } -9= { \left(x+ \sqrt{ 25- { x }^{ 2 } } \right) }^{ 2 } -2$$

Solve for x

$$x = \frac{\sqrt{\sqrt{1201} + 57}}{2} \approx 4.786842563$$

$$x = -\frac{\sqrt{57 - \sqrt{1201}}}{2} \approx -2.363501274$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\sqrt{25-x^{2}}\right)^{2}.

$$2x^{2}-9=x^{2}+2x\sqrt{25-x^{2}}+\left(\sqrt{25-x^{2}}\right)^{2}-2$$

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

$$2x^{2}-9=x^{2}+2x\sqrt{25-x^{2}}+25-x^{2}-2$$

Combine x^{2} and -x^{2} to get 0.

$$2x^{2}-9=2x\sqrt{25-x^{2}}+25-2$$

Subtract 2 from 25 to get 23.

$$2x^{2}-9=2x\sqrt{25-x^{2}}+23$$

Subtract 2x\sqrt{25-x^{2}} from both sides.

$$2x^{2}-9-2x\sqrt{25-x^{2}}=23$$

Subtract 2x^{2}-9 from both sides of the equation.

$$-2x\sqrt{25-x^{2}}=23-\left(2x^{2}-9\right)$$

To find the opposite of 2x^{2}-9, find the opposite of each term.

$$-2x\sqrt{25-x^{2}}=23-2x^{2}+9$$

Add 23 and 9 to get 32.

$$-2x\sqrt{25-x^{2}}=32-2x^{2}$$

Square both sides of the equation.

$$\left(-2x\sqrt{25-x^{2}}\right)^{2}=\left(-2x^{2}+32\right)^{2}$$

Expand \left(-2x\sqrt{25-x^{2}}\right)^{2}.

$$\left(-2\right)^{2}x^{2}\left(\sqrt{25-x^{2}}\right)^{2}=\left(-2x^{2}+32\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4x^{2}\left(\sqrt{25-x^{2}}\right)^{2}=\left(-2x^{2}+32\right)^{2}$$

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

$$4x^{2}\left(25-x^{2}\right)=\left(-2x^{2}+32\right)^{2}$$

Use the distributive property to multiply 4x^{2} by 25-x^{2}.

$$100x^{2}-4x^{4}=\left(-2x^{2}+32\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2x^{2}+32\right)^{2}.

$$100x^{2}-4x^{4}=4\left(x^{2}\right)^{2}-128x^{2}+1024$$

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

$$100x^{2}-4x^{4}=4x^{4}-128x^{2}+1024$$

Subtract 4x^{4} from both sides.

$$100x^{2}-4x^{4}-4x^{4}=-128x^{2}+1024$$

Combine -4x^{4} and -4x^{4} to get -8x^{4}.

$$100x^{2}-8x^{4}=-128x^{2}+1024$$

Add 128x^{2} to both sides.

$$100x^{2}-8x^{4}+128x^{2}=1024$$

Combine 100x^{2} and 128x^{2} to get 228x^{2}.

$$228x^{2}-8x^{4}=1024$$

Subtract 1024 from both sides.

$$228x^{2}-8x^{4}-1024=0$$

Substitute t for x^{2}.

$$-8t^{2}+228t-1024=0$$

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 -8 for a, 228 for b, and -1024 for c in the quadratic formula.

$$t=\frac{-228±\sqrt{228^{2}-4\left(-8\right)\left(-1024\right)}}{-8\times 2}$$

Do the calculations.

$$t=\frac{-228±4\sqrt{1201}}{-16}$$

Solve the equation t=\frac{-228±4\sqrt{1201}}{-16} when ± is plus and when ± is minus.

$$t=\frac{57-\sqrt{1201}}{4} t=\frac{\sqrt{1201}+57}{4}$$

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

$$x=\frac{\sqrt{57-\sqrt{1201}}}{2} x=-\frac{\sqrt{57-\sqrt{1201}}}{2} x=\frac{\sqrt{\sqrt{1201}+57}}{2} x=-\frac{\sqrt{\sqrt{1201}+57}}{2}$$

Substitute \frac{\sqrt{57-\sqrt{1201}}}{2} for x in the equation 2x^{2}-9=\left(x+\sqrt{25-x^{2}}\right)^{2}-2.

$$2\times \left(\frac{\sqrt{57-\sqrt{1201}}}{2}\right)^{2}-9=\left(\frac{\sqrt{57-\sqrt{1201}}}{2}+\sqrt{25-\left(\frac{\sqrt{57-\sqrt{1201}}}{2}\right)^{2}}\right)^{2}-2$$

Simplify. The value x=\frac{\sqrt{57-\sqrt{1201}}}{2} does not satisfy the equation.

$$\frac{39}{2}-\frac{1}{2}\times 1201^{\frac{1}{2}}=\frac{53}{2}+\frac{1}{2}\times 1201^{\frac{1}{2}}$$

Substitute -\frac{\sqrt{57-\sqrt{1201}}}{2} for x in the equation 2x^{2}-9=\left(x+\sqrt{25-x^{2}}\right)^{2}-2.

$$2\left(-\frac{\sqrt{57-\sqrt{1201}}}{2}\right)^{2}-9=\left(-\frac{\sqrt{57-\sqrt{1201}}}{2}+\sqrt{25-\left(-\frac{\sqrt{57-\sqrt{1201}}}{2}\right)^{2}}\right)^{2}-2$$

Simplify. The value x=-\frac{\sqrt{57-\sqrt{1201}}}{2} satisfies the equation.

$$\frac{39}{2}-\frac{1}{2}\times 1201^{\frac{1}{2}}=\frac{39}{2}-\frac{1}{2}\times 1201^{\frac{1}{2}}$$

Substitute \frac{\sqrt{\sqrt{1201}+57}}{2} for x in the equation 2x^{2}-9=\left(x+\sqrt{25-x^{2}}\right)^{2}-2.

$$2\times \left(\frac{\sqrt{\sqrt{1201}+57}}{2}\right)^{2}-9=\left(\frac{\sqrt{\sqrt{1201}+57}}{2}+\sqrt{25-\left(\frac{\sqrt{\sqrt{1201}+57}}{2}\right)^{2}}\right)^{2}-2$$

Simplify. The value x=\frac{\sqrt{\sqrt{1201}+57}}{2} satisfies the equation.

$$\frac{1}{2}\times 1201^{\frac{1}{2}}+\frac{39}{2}=\frac{1}{2}\times 1201^{\frac{1}{2}}+\frac{39}{2}$$

Substitute -\frac{\sqrt{\sqrt{1201}+57}}{2} for x in the equation 2x^{2}-9=\left(x+\sqrt{25-x^{2}}\right)^{2}-2.

$$2\left(-\frac{\sqrt{\sqrt{1201}+57}}{2}\right)^{2}-9=\left(-\frac{\sqrt{\sqrt{1201}+57}}{2}+\sqrt{25-\left(-\frac{\sqrt{\sqrt{1201}+57}}{2}\right)^{2}}\right)^{2}-2$$

Simplify. The value x=-\frac{\sqrt{\sqrt{1201}+57}}{2} does not satisfy the equation.

$$\frac{1}{2}\times 1201^{\frac{1}{2}}+\frac{39}{2}=-\frac{1}{2}\times 1201^{\frac{1}{2}}+\frac{53}{2}$$

List all solutions of -2x\sqrt{25-x^{2}}=32-2x^{2}.

$$x=-\frac{\sqrt{57-\sqrt{1201}}}{2} x=\frac{\sqrt{\sqrt{1201}+57}}{2}$$