Solve for x
x=-7
x=-5
Graph
Share
Copied to clipboard
a+b=12 ab=35
To solve the equation, factor x^{2}+12x+35 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
1,35 5,7
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 35.
1+35=36 5+7=12
Calculate the sum for each pair.
a=5 b=7
The solution is the pair that gives sum 12.
\left(x+5\right)\left(x+7\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=-5 x=-7
To find equation solutions, solve x+5=0 and x+7=0.
a+b=12 ab=1\times 35=35
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+35. To find a and b, set up a system to be solved.
1,35 5,7
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 35.
1+35=36 5+7=12
Calculate the sum for each pair.
a=5 b=7
The solution is the pair that gives sum 12.
\left(x^{2}+5x\right)+\left(7x+35\right)
Rewrite x^{2}+12x+35 as \left(x^{2}+5x\right)+\left(7x+35\right).
x\left(x+5\right)+7\left(x+5\right)
Factor out x in the first and 7 in the second group.
\left(x+5\right)\left(x+7\right)
Factor out common term x+5 by using distributive property.
x=-5 x=-7
To find equation solutions, solve x+5=0 and x+7=0.
x^{2}+12x+35=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{-12±\sqrt{12^{2}-4\times 35}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-12±\sqrt{144-4\times 35}}{2}
Square 12.
x=\frac{-12±\sqrt{144-140}}{2}
Multiply -4 times 35.
x=\frac{-12±\sqrt{4}}{2}
Add 144 to -140.
x=\frac{-12±2}{2}
Take the square root of 4.
x=-\frac{10}{2}
Now solve the equation x=\frac{-12±2}{2} when ± is plus. Add -12 to 2.
x=-5
Divide -10 by 2.
x=-\frac{14}{2}
Now solve the equation x=\frac{-12±2}{2} when ± is minus. Subtract 2 from -12.
x=-7
Divide -14 by 2.
x=-5 x=-7
The equation is now solved.
x^{2}+12x+35=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.
x^{2}+12x+35-35=-35
Subtract 35 from both sides of the equation.
x^{2}+12x=-35
Subtracting 35 from itself leaves 0.
x^{2}+12x+6^{2}=-35+6^{2}
Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+12x+36=-35+36
Square 6.
x^{2}+12x+36=1
Add -35 to 36.
\left(x+6\right)^{2}=1
Factor x^{2}+12x+36. 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+6\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
x+6=1 x+6=-1
Simplify.
x=-5 x=-7
Subtract 6 from both sides of the equation.
x ^ 2 +12x +35 = 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.
r + s = -12 rs = 35
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 = -6 - u s = -6 + u
Two numbers r and s sum up to -12 exactly when the average of the two numbers is \frac{1}{2}*-12 = -6. 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>
(-6 - u) (-6 + u) = 35
To solve for unknown quantity u, substitute these in the product equation rs = 35
36 - u^2 = 35
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 35-36 = -1
Simplify the expression by subtracting 36 on both sides
u^2 = 1 u = \pm\sqrt{1} = \pm 1
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-6 - 1 = -7 s = -6 + 1 = -5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
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
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}