Solve 35x^2+38x-41

$$35 x ^ { 2 } + 38 x - 41$$

Factor

$$35\left(x-\frac{-2\sqrt{449}-19}{35}\right)\left(x-\frac{2\sqrt{449}-19}{35}\right)$$

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.

$$35x^{2}+38x-41=0$$

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.

$$x=\frac{-38±\sqrt{38^{2}-4\times 35\left(-41\right)}}{2\times 35}$$

Square 38.

$$x=\frac{-38±\sqrt{1444-4\times 35\left(-41\right)}}{2\times 35}$$

Multiply -4 times 35.

$$x=\frac{-38±\sqrt{1444-140\left(-41\right)}}{2\times 35}$$

Multiply -140 times -41.

$$x=\frac{-38±\sqrt{1444+5740}}{2\times 35}$$

Add 1444 to 5740.

$$x=\frac{-38±\sqrt{7184}}{2\times 35}$$

Take the square root of 7184.

$$x=\frac{-38±4\sqrt{449}}{2\times 35}$$

Multiply 2 times 35.

$$x=\frac{-38±4\sqrt{449}}{70}$$

Now solve the equation x=\frac{-38±4\sqrt{449}}{70} when ± is plus. Add -38 to 4\sqrt{449}.

$$x=\frac{4\sqrt{449}-38}{70}$$

Divide -38+4\sqrt{449} by 70.

$$x=\frac{2\sqrt{449}-19}{35}$$

Now solve the equation x=\frac{-38±4\sqrt{449}}{70} when ± is minus. Subtract 4\sqrt{449} from -38.

$$x=\frac{-4\sqrt{449}-38}{70}$$

Divide -38-4\sqrt{449} by 70.

$$x=\frac{-2\sqrt{449}-19}{35}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-19+2\sqrt{449}}{35} for x_{1} and \frac{-19-2\sqrt{449}}{35} for x_{2}.

$$35x^{2}+38x-41=35\left(x-\frac{2\sqrt{449}-19}{35}\right)\left(x-\frac{-2\sqrt{449}-19}{35}\right)$$

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 35

$$x ^ 2 +\frac{38}{35}x -\frac{41}{35} = 0$$

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

$$r + s = -\frac{38}{35} rs = -\frac{41}{35}$$

Two numbers r and s sum up to -\frac{38}{35} exactly when the average of the two numbers is \frac{1}{2}*-\frac{38}{35} = -\frac{19}{35}. 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>

$$r = -\frac{19}{35} - u s = -\frac{19}{35} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{41}{35}

$$(-\frac{19}{35} - u) (-\frac{19}{35} + u) = -\frac{41}{35}$$

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

$$\frac{361}{1225} - u^2 = -\frac{41}{35}$$

Simplify the expression by subtracting \frac{361}{1225} on both sides

$$-u^2 = -\frac{41}{35}-\frac{361}{1225} = -\frac{1796}{1225}$$

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

$$u^2 = \frac{1796}{1225} u = \pm\sqrt{\frac{1796}{1225}} = \pm \frac{\sqrt{1796}}{35}$$

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

$$r =-\frac{19}{35} - \frac{\sqrt{1796}}{35} = -1.754 s = -\frac{19}{35} + \frac{\sqrt{1796}}{35} = 0.668$$

Evaluate

$$35x^{2}+38x-41$$