Solve 9a^2+42a-49

$$9 a ^ { 2 } + 42 a - 49$$

Factor

$$9\left(a-\frac{-7\sqrt{2}-7}{3}\right)\left(a-\frac{7\sqrt{2}-7}{3}\right)$$

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.

$$9a^{2}+42a-49=0$$

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.

$$a=\frac{-42±\sqrt{42^{2}-4\times 9\left(-49\right)}}{2\times 9}$$

Square 42.

$$a=\frac{-42±\sqrt{1764-4\times 9\left(-49\right)}}{2\times 9}$$

Multiply -4 times 9.

$$a=\frac{-42±\sqrt{1764-36\left(-49\right)}}{2\times 9}$$

Multiply -36 times -49.

$$a=\frac{-42±\sqrt{1764+1764}}{2\times 9}$$

Add 1764 to 1764.

$$a=\frac{-42±\sqrt{3528}}{2\times 9}$$

Take the square root of 3528.

$$a=\frac{-42±42\sqrt{2}}{2\times 9}$$

Multiply 2 times 9.

$$a=\frac{-42±42\sqrt{2}}{18}$$

Now solve the equation a=\frac{-42±42\sqrt{2}}{18} when ± is plus. Add -42 to 42\sqrt{2}.

$$a=\frac{42\sqrt{2}-42}{18}$$

Divide -42+42\sqrt{2} by 18.

$$a=\frac{7\sqrt{2}-7}{3}$$

Now solve the equation a=\frac{-42±42\sqrt{2}}{18} when ± is minus. Subtract 42\sqrt{2} from -42.

$$a=\frac{-42\sqrt{2}-42}{18}$$

Divide -42-42\sqrt{2} by 18.

$$a=\frac{-7\sqrt{2}-7}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-7+7\sqrt{2}}{3} for x_{1} and \frac{-7-7\sqrt{2}}{3} for x_{2}.

$$9a^{2}+42a-49=9\left(a-\frac{7\sqrt{2}-7}{3}\right)\left(a-\frac{-7\sqrt{2}-7}{3}\right)$$

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

$$x ^ 2 +\frac{14}{3}x -\frac{49}{9} = 0$$

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

$$r + s = -\frac{14}{3} rs = -\frac{49}{9}$$

Two numbers r and s sum up to -\frac{14}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{14}{3} = -\frac{7}{3}. 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>

$$r = -\frac{7}{3} - u s = -\frac{7}{3} + u$$

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

$$(-\frac{7}{3} - u) (-\frac{7}{3} + u) = -\frac{49}{9}$$

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

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

Simplify the expression by subtracting \frac{49}{9} on both sides

$$-u^2 = -\frac{49}{9}-\frac{49}{9} = -\frac{98}{9}$$

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

$$u^2 = \frac{98}{9} u = \pm\sqrt{\frac{98}{9}} = \pm \frac{\sqrt{98}}{3}$$

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

$$r =-\frac{7}{3} - \frac{\sqrt{98}}{3} = -5.633 s = -\frac{7}{3} + \frac{\sqrt{98}}{3} = 0.966$$

Evaluate

$$9a^{2}+42a-49$$