Solve 10x^2+49x+10

$$10 x ^ { 2 } + 49 x + 10$$

Factor

$$10\left(x-\frac{-\sqrt{2001}-49}{20}\right)\left(x-\frac{\sqrt{2001}-49}{20}\right)$$

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.

$$10x^{2}+49x+10=0$$

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.

$$x=\frac{-49±\sqrt{49^{2}-4\times 10\times 10}}{2\times 10}$$

Square 49.

$$x=\frac{-49±\sqrt{2401-4\times 10\times 10}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-49±\sqrt{2401-40\times 10}}{2\times 10}$$

Multiply -40 times 10.

$$x=\frac{-49±\sqrt{2401-400}}{2\times 10}$$

Add 2401 to -400.

$$x=\frac{-49±\sqrt{2001}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-49±\sqrt{2001}}{20}$$

Now solve the equation x=\frac{-49±\sqrt{2001}}{20} when ± is plus. Add -49 to \sqrt{2001}.

$$x=\frac{\sqrt{2001}-49}{20}$$

Now solve the equation x=\frac{-49±\sqrt{2001}}{20} when ± is minus. Subtract \sqrt{2001} from -49.

$$x=\frac{-\sqrt{2001}-49}{20}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-49+\sqrt{2001}}{20} for x_{1} and \frac{-49-\sqrt{2001}}{20} for x_{2}.

$$10x^{2}+49x+10=10\left(x-\frac{\sqrt{2001}-49}{20}\right)\left(x-\frac{-\sqrt{2001}-49}{20}\right)$$

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.This is achieved by dividing both sides of the equation by 10

$$x ^ 2 +\frac{49}{10}x +1 = 0$$

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

$$r + s = -\frac{49}{10} rs = 1$$

Two numbers r and s sum up to -\frac{49}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{49}{10} = -\frac{49}{20}. 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>

$$r = -\frac{49}{20} - u s = -\frac{49}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(-\frac{49}{20} - u) (-\frac{49}{20} + u) = 1$$

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

$$\frac{2401}{400} - u^2 = 1$$

Simplify the expression by subtracting \frac{2401}{400} on both sides

$$-u^2 = 1-\frac{2401}{400} = -\frac{2001}{400}$$

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

$$u^2 = \frac{2001}{400} u = \pm\sqrt{\frac{2001}{400}} = \pm \frac{\sqrt{2001}}{20}$$

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

$$r =-\frac{49}{20} - \frac{\sqrt{2001}}{20} = -4.687 s = -\frac{49}{20} + \frac{\sqrt{2001}}{20} = -0.213$$

Evaluate

$$10x^{2}+49x+10$$