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a+b=-7 ab=12
To solve the equation, factor x^{2}-7x+12 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,-12 -2,-6 -3,-4
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 12.
-1-12=-13 -2-6=-8 -3-4=-7
Calculate the sum for each pair.
a=-4 b=-3
The solution is the pair that gives sum -7.
\left(x-4\right)\left(x-3\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=4 x=3
To find equation solutions, solve x-4=0 and x-3=0.
a+b=-7 ab=1\times 12=12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.
-1,-12 -2,-6 -3,-4
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 12.
-1-12=-13 -2-6=-8 -3-4=-7
Calculate the sum for each pair.
a=-4 b=-3
The solution is the pair that gives sum -7.
\left(x^{2}-4x\right)+\left(-3x+12\right)
Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).
x\left(x-4\right)-3\left(x-4\right)
Factor out x in the first and -3 in the second group.
\left(x-4\right)\left(x-3\right)
Factor out common term x-4 by using distributive property.
x=4 x=3
To find equation solutions, solve x-4=0 and x-3=0.
x^{2}-7x+12=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-48}}{2}
Multiply -4 times 12.
x=\frac{-\left(-7\right)±\sqrt{1}}{2}
Add 49 to -48.
x=\frac{-\left(-7\right)±1}{2}
Take the square root of 1.
x=\frac{7±1}{2}
The opposite of -7 is 7.
x=\frac{8}{2}
Now solve the equation x=\frac{7±1}{2} when ± is plus. Add 7 to 1.
x=4
Divide 8 by 2.
x=\frac{6}{2}
Now solve the equation x=\frac{7±1}{2} when ± is minus. Subtract 1 from 7.
x=3
Divide 6 by 2.
x=4 x=3
The equation is now solved.
x^{2}-7x+12=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}-7x+12-12=-12
Subtract 12 from both sides of the equation.
x^{2}-7x=-12
Subtracting 12 from itself leaves 0.
x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}
Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}
Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-7x+\frac{49}{4}=\frac{1}{4}
Add -12 to \frac{49}{4}.
\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}
Factor x^{2}-7x+\frac{49}{4}. 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-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}
Simplify.
x=4 x=3
Add \frac{7}{2} to both sides of the equation.
x ^ 2 -7x +12 = 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 = 7 rs = 12
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 = \frac{7}{2} - u s = \frac{7}{2} + u
Two numbers r and s sum up to 7 exactly when the average of the two numbers is \frac{1}{2}*7 = \frac{7}{2}. 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>
(\frac{7}{2} - u) (\frac{7}{2} + u) = 12
To solve for unknown quantity u, substitute these in the product equation rs = 12
\frac{49}{4} - u^2 = 12
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 12-\frac{49}{4} = -\frac{1}{4}
Simplify the expression by subtracting \frac{49}{4} on both sides
u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{2} - \frac{1}{2} = 3 s = \frac{7}{2} + \frac{1}{2} = 4
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.