Solve -x^2-6x-9=

$$- x ^ { 2 } - 6 x - 9 =$$

Factor

$$-\left(x+3\right)^{2}$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-6 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

$$a=-3 b=-3$$

Rewrite -x^{2}-6x-9 as \left(-x^{2}-3x\right)+\left(-3x-9\right).

$$\left(-x^{2}-3x\right)+\left(-3x-9\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x+3\right)-3\left(x+3\right)$$

Factor out common term x+3 by using distributive property.

$$\left(x+3\right)\left(-x-3\right)$$

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.

$$-x^{2}-6x-9=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-6\right)±\sqrt{36+4\left(-9\right)}}{2\left(-1\right)}$$

Multiply 4 times -9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2\left(-1\right)}$$

Add 36 to -36.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$x=\frac{-\left(-6\right)±0}{2\left(-1\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±0}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{6±0}{-2}$$

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

$$-x^{2}-6x-9=-\left(x-\left(-3\right)\right)\left(x-\left(-3\right)\right)$$

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

$$-x^{2}-6x-9=-\left(x+3\right)\left(x+3\right)$$

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.

$$x ^ 2 +6x +9 = 0$$

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

$$r + s = -6 rs = 9$$

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

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 9

$$(-3 - u) (-3 + u) = 9$$

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

$$9 - u^2 = 9$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 9-9 = 0$$

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

$$u^2 = 0 u = 0$$

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

$$r = s = -3$$

Evaluate

$$-\left(x+3\right)^{2}$$