Solve x-sqrt{25-x^2}=1

$$x - \sqrt { 25 - x ^ { 2 } } = 1$$

Solve for x

$$x=4$$

Solution Steps

Subtract x from both sides of the equation.

$$-\sqrt{25-x^{2}}=1-x$$

Square both sides of the equation.

$$\left(-\sqrt{25-x^{2}}\right)^{2}=\left(1-x\right)^{2}$$

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

$$\left(-1\right)^{2}\left(\sqrt{25-x^{2}}\right)^{2}=\left(1-x\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$1\left(\sqrt{25-x^{2}}\right)^{2}=\left(1-x\right)^{2}$$

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

$$1\left(25-x^{2}\right)=\left(1-x\right)^{2}$$

Use the distributive property to multiply 1 by 25-x^{2}.

$$25-x^{2}=\left(1-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$25-x^{2}=1-2x+x^{2}$$

Subtract 1 from both sides.

$$25-x^{2}-1=-2x+x^{2}$$

Subtract 1 from 25 to get 24.

$$24-x^{2}=-2x+x^{2}$$

Add 2x to both sides.

$$24-x^{2}+2x=x^{2}$$

Subtract x^{2} from both sides.

$$24-x^{2}+2x-x^{2}=0$$

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

$$24-2x^{2}+2x=0$$

Divide both sides by 2.

$$12-x^{2}+x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=1 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -x^{2}+x+12 as \left(-x^{2}+4x\right)+\left(-3x+12\right).

$$\left(-x^{2}+4x\right)+\left(-3x+12\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x-4\right)-3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-3\right)$$

To find equation solutions, solve x-4=0 and -x-3=0.

$$x=4 x=-3$$

Substitute 4 for x in the equation x-\sqrt{25-x^{2}}=1.

$$4-\sqrt{25-4^{2}}=1$$

Simplify. The value x=4 satisfies the equation.

$$1=1$$

Substitute -3 for x in the equation x-\sqrt{25-x^{2}}=1.

$$-3-\sqrt{25-\left(-3\right)^{2}}=1$$

Simplify. The value x=-3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-7=1$$

Equation -\sqrt{25-x^{2}}=1-x has a unique solution.

$$x=4$$