Solve (4-3x)^2geq-2

$$( 4 - 3 x ) ^ { 2 } \geq - 2$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$\left(4-3x\right)^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 9 for a, -24 for b, and 18 for c in the quadratic formula.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 9\times 18}}{2\times 9}$$

Do the calculations.

$$x=\frac{24±\sqrt{-72}}{18}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression \left(4-3x\right)^{2} has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$\left(4-3\times 0\right)^{2}=16$$

The value of the expression \left(4-3x\right)^{2} is always ≥0. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$