Solve 1-4x^2+x+9

$$1-4 { x }^{ 2 } +x+9$$

Factor

$$-4\left(x-\frac{1-\sqrt{161}}{8}\right)\left(x-\frac{\sqrt{161}+1}{8}\right)$$

Steps Using the Quadratic Formula

Add 1 and 9 to get 10.

$$factor(10-4x^{2}+x)$$

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.

$$-4x^{2}+x+10=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-4\right)\times 10}}{2\left(-4\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-4\right)\times 10}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-1±\sqrt{1+16\times 10}}{2\left(-4\right)}$$

Multiply 16 times 10.

$$x=\frac{-1±\sqrt{1+160}}{2\left(-4\right)}$$

Add 1 to 160.

$$x=\frac{-1±\sqrt{161}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-1±\sqrt{161}}{-8}$$

Now solve the equation x=\frac{-1±\sqrt{161}}{-8} when ± is plus. Add -1 to \sqrt{161}.

$$x=\frac{\sqrt{161}-1}{-8}$$

Divide -1+\sqrt{161} by -8.

$$x=\frac{1-\sqrt{161}}{8}$$

Now solve the equation x=\frac{-1±\sqrt{161}}{-8} when ± is minus. Subtract \sqrt{161} from -1.

$$x=\frac{-\sqrt{161}-1}{-8}$$

Divide -1-\sqrt{161} by -8.

$$x=\frac{\sqrt{161}+1}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1-\sqrt{161}}{8} for x_{1} and \frac{1+\sqrt{161}}{8} for x_{2}.

$$-4x^{2}+x+10=-4\left(x-\frac{1-\sqrt{161}}{8}\right)\left(x-\frac{\sqrt{161}+1}{8}\right)$$

Evaluate

$$10+x-4x^{2}$$

Solution Steps

Add 1 and 9 to get 10.

$$10-4x^{2}+x$$