Solve 4(x^2-49)+61geq(2x-5)^2

$$4 ( x ^ { 2 } - 49 ) + 61 \geq ( 2 x - 5 ) ^ { 2 }$$

Solve for x

$$x\geq 8$$

Solution Steps

Use the distributive property to multiply 4 by x^{2}-49.

$$4x^{2}-196+61\geq \left(2x-5\right)^{2}$$

Add -196 and 61 to get -135.

$$4x^{2}-135\geq \left(2x-5\right)^{2}$$

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

$$4x^{2}-135\geq 4x^{2}-20x+25$$

Subtract 4x^{2} from both sides.

$$4x^{2}-135-4x^{2}\geq -20x+25$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-135\geq -20x+25$$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$-20x+25\leq -135$$

Subtract 25 from both sides.

$$-20x\leq -135-25$$

Subtract 25 from -135 to get -160.

$$-20x\leq -160$$

Divide both sides by -20. Since -20 is negative, the inequality direction is changed.

$$x\geq \frac{-160}{-20}$$

Divide -160 by -20 to get 8.

$$x\geq 8$$