Solve 4xleft(x-2right)leq11+left(x-4right)^2

$$4x \left( x-2 \right) \leq 11+ { \left(x-4 \right) }^{ 2 }$$

Solve for x

$$x\in \begin{bmatrix}-3,3\end{bmatrix}$$

Solution Steps

Use the distributive property to multiply 4x by x-2.

$$4x^{2}-8x\leq 11+\left(x-4\right)^{2}$$

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

$$4x^{2}-8x\leq 11+x^{2}-8x+16$$

Add 11 and 16 to get 27.

$$4x^{2}-8x\leq 27+x^{2}-8x$$

Subtract x^{2} from both sides.

$$4x^{2}-8x-x^{2}\leq 27-8x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-8x\leq 27-8x$$

Add 8x to both sides.

$$3x^{2}-8x+8x\leq 27$$

Combine -8x and 8x to get 0.

$$3x^{2}\leq 27$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x^{2}\leq \frac{27}{3}$$

Divide 27 by 3 to get 9.

$$x^{2}\leq 9$$

Calculate the square root of 9 and get 3. Rewrite 9 as 3^{2}.

$$x^{2}\leq 3^{2}$$

Inequality holds for |x|\leq 3.

$$|x|\leq 3$$

Rewrite |x|\leq 3 as x\in \left[-3,3\right].

$$x\in \begin{bmatrix}-3,3\end{bmatrix}$$