Solve 4x^2-35/4x+1geq0

$$4 x ^ { 2 } - \frac { 35 } { 4 } x + 1 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{35-\sqrt{969}}{32}]\cup [\frac{\sqrt{969}+35}{32},\infty)$$

Steps Using the Quadratic Formula

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.

$$4x^{2}-\frac{35}{4}x+1=0$$

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 4 for a, -\frac{35}{4} for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-\frac{35}{4}\right)±\sqrt{\left(-\frac{35}{4}\right)^{2}-4\times 4\times 1}}{2\times 4}$$

Do the calculations.

$$x=\frac{\frac{35}{4}±\frac{1}{4}\sqrt{969}}{8}$$

Solve the equation x=\frac{\frac{35}{4}±\frac{1}{4}\sqrt{969}}{8} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{969}+35}{32} x=\frac{35-\sqrt{969}}{32}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{\sqrt{969}+35}{32}\right)\left(x-\frac{35-\sqrt{969}}{32}\right)\geq 0$$

For the product to be ≥0, x-\frac{\sqrt{969}+35}{32} and x-\frac{35-\sqrt{969}}{32} have to be both ≤0 or both ≥0. Consider the case when x-\frac{\sqrt{969}+35}{32} and x-\frac{35-\sqrt{969}}{32} are both ≤0.

$$x-\frac{\sqrt{969}+35}{32}\leq 0 x-\frac{35-\sqrt{969}}{32}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{35-\sqrt{969}}{32}.

$$x\leq \frac{35-\sqrt{969}}{32}$$

Consider the case when x-\frac{\sqrt{969}+35}{32} and x-\frac{35-\sqrt{969}}{32} are both ≥0.

$$x-\frac{35-\sqrt{969}}{32}\geq 0 x-\frac{\sqrt{969}+35}{32}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{969}+35}{32}.

$$x\geq \frac{\sqrt{969}+35}{32}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{35-\sqrt{969}}{32}\text{; }x\geq \frac{\sqrt{969}+35}{32}$$