Solve 49a^2+42a+4

$$49 a ^ { 2 } + 42 a + 4$$

Factor

$$49\left(a-\frac{-\sqrt{5}-3}{7}\right)\left(a-\frac{\sqrt{5}-3}{7}\right)$$

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.

$$49a^{2}+42a+4=0$$

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.

$$a=\frac{-42±\sqrt{42^{2}-4\times 49\times 4}}{2\times 49}$$

Square 42.

$$a=\frac{-42±\sqrt{1764-4\times 49\times 4}}{2\times 49}$$

Multiply -4 times 49.

$$a=\frac{-42±\sqrt{1764-196\times 4}}{2\times 49}$$

Multiply -196 times 4.

$$a=\frac{-42±\sqrt{1764-784}}{2\times 49}$$

Add 1764 to -784.

$$a=\frac{-42±\sqrt{980}}{2\times 49}$$

Take the square root of 980.

$$a=\frac{-42±14\sqrt{5}}{2\times 49}$$

Multiply 2 times 49.

$$a=\frac{-42±14\sqrt{5}}{98}$$

Now solve the equation a=\frac{-42±14\sqrt{5}}{98} when ± is plus. Add -42 to 14\sqrt{5}.

$$a=\frac{14\sqrt{5}-42}{98}$$

Divide -42+14\sqrt{5} by 98.

$$a=\frac{\sqrt{5}-3}{7}$$

Now solve the equation a=\frac{-42±14\sqrt{5}}{98} when ± is minus. Subtract 14\sqrt{5} from -42.

$$a=\frac{-14\sqrt{5}-42}{98}$$

Divide -42-14\sqrt{5} by 98.

$$a=\frac{-\sqrt{5}-3}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{5}}{7} for x_{1} and \frac{-3-\sqrt{5}}{7} for x_{2}.

$$49a^{2}+42a+4=49\left(a-\frac{\sqrt{5}-3}{7}\right)\left(a-\frac{-\sqrt{5}-3}{7}\right)$$

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 49

$$x ^ 2 +\frac{6}{7}x +\frac{4}{49} = 0$$

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

$$r + s = -\frac{6}{7} rs = \frac{4}{49}$$

Two numbers r and s sum up to -\frac{6}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{6}{7} = -\frac{3}{7}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{49}

$$(-\frac{3}{7} - u) (-\frac{3}{7} + u) = \frac{4}{49}$$

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

$$\frac{9}{49} - u^2 = \frac{4}{49}$$

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

$$-u^2 = \frac{4}{49}-\frac{9}{49} = -\frac{5}{49}$$

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

$$u^2 = \frac{5}{49} u = \pm\sqrt{\frac{5}{49}} = \pm \frac{\sqrt{5}}{7}$$

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

$$r =-\frac{3}{7} - \frac{\sqrt{5}}{7} = -0.748 s = -\frac{3}{7} + \frac{\sqrt{5}}{7} = -0.109$$

Evaluate

$$49a^{2}+42a+4$$