Solve 9x^2+53x+49

$$9 x ^ { 2 } + 53 x + 49$$

Factor

$$9\left(x-\frac{-\sqrt{1045}-53}{18}\right)\left(x-\frac{\sqrt{1045}-53}{18}\right)$$

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.

$$9x^{2}+53x+49=0$$

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.

$$x=\frac{-53±\sqrt{53^{2}-4\times 9\times 49}}{2\times 9}$$

Square 53.

$$x=\frac{-53±\sqrt{2809-4\times 9\times 49}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-53±\sqrt{2809-36\times 49}}{2\times 9}$$

Multiply -36 times 49.

$$x=\frac{-53±\sqrt{2809-1764}}{2\times 9}$$

Add 2809 to -1764.

$$x=\frac{-53±\sqrt{1045}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-53±\sqrt{1045}}{18}$$

Now solve the equation x=\frac{-53±\sqrt{1045}}{18} when ± is plus. Add -53 to \sqrt{1045}.

$$x=\frac{\sqrt{1045}-53}{18}$$

Now solve the equation x=\frac{-53±\sqrt{1045}}{18} when ± is minus. Subtract \sqrt{1045} from -53.

$$x=\frac{-\sqrt{1045}-53}{18}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-53+\sqrt{1045}}{18} for x_{1} and \frac{-53-\sqrt{1045}}{18} for x_{2}.

$$9x^{2}+53x+49=9\left(x-\frac{\sqrt{1045}-53}{18}\right)\left(x-\frac{-\sqrt{1045}-53}{18}\right)$$

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 9

$$x ^ 2 +\frac{53}{9}x +\frac{49}{9} = 0$$

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

$$r + s = -\frac{53}{9} rs = \frac{49}{9}$$

Two numbers r and s sum up to -\frac{53}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{53}{9} = -\frac{53}{18}. 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>

$$r = -\frac{53}{18} - u s = -\frac{53}{18} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{49}{9}

$$(-\frac{53}{18} - u) (-\frac{53}{18} + u) = \frac{49}{9}$$

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

$$\frac{2809}{324} - u^2 = \frac{49}{9}$$

Simplify the expression by subtracting \frac{2809}{324} on both sides

$$-u^2 = \frac{49}{9}-\frac{2809}{324} = -\frac{1045}{324}$$

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

$$u^2 = \frac{1045}{324} u = \pm\sqrt{\frac{1045}{324}} = \pm \frac{\sqrt{1045}}{18}$$

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

$$r =-\frac{53}{18} - \frac{\sqrt{1045}}{18} = -4.740 s = -\frac{53}{18} + \frac{\sqrt{1045}}{18} = -1.149$$

Evaluate

$$9x^{2}+53x+49$$