Solve f(x)=-4x^2+6x+1

$$f ( x ) = - 4 x ^ { 2 } + 6 x + 1$$

Factor

$$-4\left(x-\frac{3-\sqrt{13}}{4}\right)\left(x-\frac{\sqrt{13}+3}{4}\right)$$

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.

$$-4x^{2}+6x+1=0$$

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

$$x=\frac{-6±\sqrt{6^{2}-4\left(-4\right)}}{2\left(-4\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-4\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-6±\sqrt{36+16}}{2\left(-4\right)}$$

Add 36 to 16.

$$x=\frac{-6±\sqrt{52}}{2\left(-4\right)}$$

Take the square root of 52.

$$x=\frac{-6±2\sqrt{13}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-6±2\sqrt{13}}{-8}$$

Now solve the equation x=\frac{-6±2\sqrt{13}}{-8} when ± is plus. Add -6 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}-6}{-8}$$

Divide -6+2\sqrt{13} by -8.

$$x=\frac{3-\sqrt{13}}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{13}}{-8} when ± is minus. Subtract 2\sqrt{13} from -6.

$$x=\frac{-2\sqrt{13}-6}{-8}$$

Divide -6-2\sqrt{13} by -8.

$$x=\frac{\sqrt{13}+3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3-\sqrt{13}}{4} for x_{1} and \frac{3+\sqrt{13}}{4} for x_{2}.

$$-4x^{2}+6x+1=-4\left(x-\frac{3-\sqrt{13}}{4}\right)\left(x-\frac{\sqrt{13}+3}{4}\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -\frac{3}{2}x -\frac{1}{4} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{3}{2} rs = -\frac{1}{4}$$

Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{4}

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = -\frac{1}{4}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{16} - u^2 = -\frac{1}{4}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -\frac{1}{4}-\frac{9}{16} = -\frac{13}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{13}{16} u = \pm\sqrt{\frac{13}{16}} = \pm \frac{\sqrt{13}}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{3}{4} - \frac{\sqrt{13}}{4} = -0.151 s = \frac{3}{4} + \frac{\sqrt{13}}{4} = 1.651$$

Evaluate

$$1+6x-4x^{2}$$