Solve 2x^3-7x+5=0

$$2 x ^ { 3 } - 7 x + 5 = 0$$

Solve for x

$$x = \frac{\sqrt{11} - 1}{2} \approx 1.158312395$$

$$x=\frac{-\sqrt{11}-1}{2}\approx -2.158312395$$

$$x=1$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-7x+5 by x-1 to get 2x^{2}+2x-5. Solve the equation where the result equals to 0.

$$2x^{2}+2x-5=0$$

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 2 for a, 2 for b, and -5 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-2±2\sqrt{11}}{4}$$

Solve the equation 2x^{2}+2x-5=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{11}-1}{2} x=\frac{\sqrt{11}-1}{2}$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{11}-1}{2} x=\frac{\sqrt{11}-1}{2}$$