Solve 3x+4sqrt{1-x^2}=5

$$3 x + 4 \sqrt { 1 - x ^ { 2 } } = 5$$

Solve for x

$$x=\frac{3}{5}=0.6$$

Solution Steps

Subtract 3x from both sides of the equation.

$$4\sqrt{1-x^{2}}=5-3x$$

Square both sides of the equation.

$$\left(4\sqrt{1-x^{2}}\right)^{2}=\left(5-3x\right)^{2}$$

Expand \left(4\sqrt{1-x^{2}}\right)^{2}.

$$4^{2}\left(\sqrt{1-x^{2}}\right)^{2}=\left(5-3x\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16\left(\sqrt{1-x^{2}}\right)^{2}=\left(5-3x\right)^{2}$$

Calculate \sqrt{1-x^{2}} to the power of 2 and get 1-x^{2}.

$$16\left(1-x^{2}\right)=\left(5-3x\right)^{2}$$

Use the distributive property to multiply 16 by 1-x^{2}.

$$16-16x^{2}=\left(5-3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-3x\right)^{2}.

$$16-16x^{2}=25-30x+9x^{2}$$

Subtract 25 from both sides.

$$16-16x^{2}-25=-30x+9x^{2}$$

Subtract 25 from 16 to get -9.

$$-9-16x^{2}=-30x+9x^{2}$$

Add 30x to both sides.

$$-9-16x^{2}+30x=9x^{2}$$

Subtract 9x^{2} from both sides.

$$-9-16x^{2}+30x-9x^{2}=0$$

Combine -16x^{2} and -9x^{2} to get -25x^{2}.

$$-9-25x^{2}+30x=0$$

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

$$-25x^{2}+30x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -25x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=30 ab=-25\left(-9\right)=225$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 225.

$$1,225 3,75 5,45 9,25 15,15$$

Calculate the sum for each pair.

$$1+225=226 3+75=78 5+45=50 9+25=34 15+15=30$$

The solution is the pair that gives sum 30.

$$a=15 b=15$$

Rewrite -25x^{2}+30x-9 as \left(-25x^{2}+15x\right)+\left(15x-9\right).

$$\left(-25x^{2}+15x\right)+\left(15x-9\right)$$

Factor out -5x in the first and 3 in the second group.

$$-5x\left(5x-3\right)+3\left(5x-3\right)$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(-5x+3\right)$$

To find equation solutions, solve 5x-3=0 and -5x+3=0.

$$x=\frac{3}{5} x=\frac{3}{5}$$

Substitute \frac{3}{5} for x in the equation 3x+4\sqrt{1-x^{2}}=5.

$$3\times \frac{3}{5}+4\sqrt{1-\left(\frac{3}{5}\right)^{2}}=5$$

Simplify. The value x=\frac{3}{5} satisfies the equation.

$$5=5$$

Substitute \frac{3}{5} for x in the equation 3x+4\sqrt{1-x^{2}}=5.

$$3\times \frac{3}{5}+4\sqrt{1-\left(\frac{3}{5}\right)^{2}}=5$$

Simplify. The value x=\frac{3}{5} satisfies the equation.

$$5=5$$

List all solutions of 4\sqrt{1-x^{2}}=5-3x.

$$x=\frac{3}{5} x=\frac{3}{5}$$