Solve for x
x=-3
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x^{2}+6x+9=0
Divide both sides by 3.
a+b=6 ab=1\times 9=9
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+9. To find a and b, set up a system to be solved.
1,9 3,3
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.
1+9=10 3+3=6
Calculate the sum for each pair.
a=3 b=3
The solution is the pair that gives sum 6.
\left(x^{2}+3x\right)+\left(3x+9\right)
Rewrite x^{2}+6x+9 as \left(x^{2}+3x\right)+\left(3x+9\right).
x\left(x+3\right)+3\left(x+3\right)
Factor out x in the first and 3 in the second group.
\left(x+3\right)\left(x+3\right)
Factor out common term x+3 by using distributive property.
\left(x+3\right)^{2}
Rewrite as a binomial square.
x=-3
To find equation solution, solve x+3=0.
3x^{2}+18x+27=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{-18±\sqrt{18^{2}-4\times 3\times 27}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 18 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-18±\sqrt{324-4\times 3\times 27}}{2\times 3}
Square 18.
x=\frac{-18±\sqrt{324-12\times 27}}{2\times 3}
Multiply -4 times 3.
x=\frac{-18±\sqrt{324-324}}{2\times 3}
Multiply -12 times 27.
x=\frac{-18±\sqrt{0}}{2\times 3}
Add 324 to -324.
x=-\frac{18}{2\times 3}
Take the square root of 0.
x=-\frac{18}{6}
Multiply 2 times 3.
x=-3
Divide -18 by 6.
3x^{2}+18x+27=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
3x^{2}+18x+27-27=-27
Subtract 27 from both sides of the equation.
3x^{2}+18x=-27
Subtracting 27 from itself leaves 0.
\frac{3x^{2}+18x}{3}=-\frac{27}{3}
Divide both sides by 3.
x^{2}+\frac{18}{3}x=-\frac{27}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+6x=-\frac{27}{3}
Divide 18 by 3.
x^{2}+6x=-9
Divide -27 by 3.
x^{2}+6x+3^{2}=-9+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+6x+9=-9+9
Square 3.
x^{2}+6x+9=0
Add -9 to 9.
\left(x+3\right)^{2}=0
Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+3\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x+3=0 x+3=0
Simplify.
x=-3 x=-3
Subtract 3 from both sides of the equation.
x=-3
The equation is now solved. Solutions are the same.
x ^ 2 +6x +9 = 0
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 3
r + s = -6 rs = 9
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 = -3 - u s = -3 + u
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>
(-3 - u) (-3 + u) = 9
To solve for unknown quantity u, substitute these in the product equation rs = 9
9 - u^2 = 9
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 9-9 = 0
Simplify the expression by subtracting 9 on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = -3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
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699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
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