Solve (x-5)^2leq(x-7)*(x-4)

$$( x - 5 ) ^ { 2 } \leq ( x - 7 ) \cdot ( x - 4 )$$

Solve for x

$$x\leq 3$$

Solution Steps

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

$$x^{2}-10x+25\leq \left(x-7\right)\left(x-4\right)$$

Use the distributive property to multiply x-7 by x-4 and combine like terms.

$$x^{2}-10x+25\leq x^{2}-11x+28$$

Subtract x^{2} from both sides.

$$x^{2}-10x+25-x^{2}\leq -11x+28$$

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

$$-10x+25\leq -11x+28$$

Add 11x to both sides.

$$-10x+25+11x\leq 28$$

Combine -10x and 11x to get x.

$$x+25\leq 28$$

Subtract 25 from both sides.

$$x\leq 28-25$$

Subtract 25 from 28 to get 3.

$$x\leq 3$$