Solve x^2+28x-29=

$$x ^ { 2 } + 28 x - 29 =$$

Factor

$$\left(x-1\right)\left(x+29\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-29. To find a and b, set up a system to be solved.

$$a+b=28 ab=1\left(-29\right)=-29$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=29$$

Rewrite x^{2}+28x-29 as \left(x^{2}-x\right)+\left(29x-29\right).

$$\left(x^{2}-x\right)+\left(29x-29\right)$$

Factor out x in the first and 29 in the second group.

$$x\left(x-1\right)+29\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+29\right)$$

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.

$$x^{2}+28x-29=0$$

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.

$$x=\frac{-28±\sqrt{28^{2}-4\left(-29\right)}}{2}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\left(-29\right)}}{2}$$

Multiply -4 times -29.

$$x=\frac{-28±\sqrt{784+116}}{2}$$

Add 784 to 116.

$$x=\frac{-28±\sqrt{900}}{2}$$

Take the square root of 900.

$$x=\frac{-28±30}{2}$$

Now solve the equation x=\frac{-28±30}{2} when ± is plus. Add -28 to 30.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-28±30}{2} when ± is minus. Subtract 30 from -28.

$$x=-\frac{58}{2}$$

Divide -58 by 2.

$$x=-29$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -29 for x_{2}.

$$x^{2}+28x-29=\left(x-1\right)\left(x-\left(-29\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}+28x-29=\left(x-1\right)\left(x+29\right)$$

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.

$$x ^ 2 +28x -29 = 0$$

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

$$r + s = -28 rs = -29$$

Two numbers r and s sum up to -28 exactly when the average of the two numbers is \frac{1}{2}*-28 = -14. 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>

$$r = -14 - u s = -14 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -29

$$(-14 - u) (-14 + u) = -29$$

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

$$196 - u^2 = -29$$

Simplify the expression by subtracting 196 on both sides

$$-u^2 = -29-196 = -225$$

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

$$u^2 = 225 u = \pm\sqrt{225} = \pm 15$$

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

$$r =-14 - 15 = -29 s = -14 + 15 = 1$$

Evaluate

$$\left(x-1\right)\left(x+29\right)$$