Solve {left(sqrt{x^2+18x+17}right)}^2<left(-x+3right)^2

$${ \left( \sqrt{ { x }^{ 2 } +18x+17 } \right) }^{ 2 } < { \left(-x+3 \right) }^{ 2 }$$

Solve for x

$$x\in (-\infty,-17]\cup [-1,-\frac{1}{3})$$

Solution Steps

Calculate \sqrt{x^{2}+18x+17} to the power of 2 and get x^{2}+18x+17.

$$x^{2}+18x+17<\left(-x+3\right)^{2}$$

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

$$x^{2}+18x+17<\left(-x\right)^{2}+6\left(-x\right)+9$$

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

$$x^{2}+18x+17<x^{2}+6\left(-x\right)+9$$

Subtract x^{2} from both sides.

$$x^{2}+18x+17-x^{2}<6\left(-x\right)+9$$

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

$$18x+17<6\left(-x\right)+9$$

Subtract 6\left(-x\right) from both sides.

$$18x+17-6\left(-x\right)<9$$

Multiply -1 and 6 to get -6.

$$18x+17-6\left(-1\right)x<9$$

Multiply -6 and -1 to get 6.

$$18x+17+6x<9$$

Combine 18x and 6x to get 24x.

$$24x+17<9$$

Subtract 17 from both sides.

$$24x<9-17$$

Subtract 17 from 9 to get -8.

$$24x<-8$$

Divide both sides by 24. Since 24 is positive, the inequality direction remains the same.

$$x<\frac{-8}{24}$$

Reduce the fraction \frac{-8}{24} to lowest terms by extracting and canceling out 8.

$$x<-\frac{1}{3}$$