Solve 2x^2+4x+98

$$2 { x }^{ 2 } +4x+98$$

Factor

$$2\left(x^{2}+2x+49\right)$$

Solution Steps

Factor out 2. Polynomial x^{2}+2x+49 is not factored since it does not have any rational roots.

$$2\left(x^{2}+2x+49\right)$$

Steps Using the Quadratic Formula

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.

$$2x^{2}+4x+98=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\times 98}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\times 98}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\times 98}}{2\times 2}$$

Multiply -8 times 98.

$$x=\frac{-4±\sqrt{16-784}}{2\times 2}$$

Add 16 to -784.

$$x=\frac{-4±\sqrt{-768}}{2\times 2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$2x^{2}+4x+98$$

Evaluate

$$2\left(x^{2}+2x+49\right)$$