Solve 4(x^2+1)+8(3x-4)>4x^2

$$4 ( x ^ { 2 } + 1 ) + 8 ( 3 x - 4 ) > 4 x ^ { 2 }$$

Solve for x

$$x>\frac{7}{6}$$

Solution Steps

Use the distributive property to multiply 4 by x^{2}+1.

$$4x^{2}+4+8\left(3x-4\right)>4x^{2}$$

Use the distributive property to multiply 8 by 3x-4.

$$4x^{2}+4+24x-32>4x^{2}$$

Subtract 32 from 4 to get -28.

$$4x^{2}-28+24x>4x^{2}$$

Subtract 4x^{2} from both sides.

$$4x^{2}-28+24x-4x^{2}>0$$

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

$$-28+24x>0$$

Add 28 to both sides. Anything plus zero gives itself.

$$24x>28$$

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

$$x>\frac{28}{24}$$

Reduce the fraction \frac{28}{24} to lowest terms by extracting and canceling out 4.

$$x>\frac{7}{6}$$