Solve for k
k=-10
k=-3
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a+b=13 ab=30
To solve the equation, factor k^{2}+13k+30 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.
1,30 2,15 3,10 5,6
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 30.
1+30=31 2+15=17 3+10=13 5+6=11
Calculate the sum for each pair.
a=3 b=10
The solution is the pair that gives sum 13.
\left(k+3\right)\left(k+10\right)
Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.
k=-3 k=-10
To find equation solutions, solve k+3=0 and k+10=0.
a+b=13 ab=1\times 30=30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+30. To find a and b, set up a system to be solved.
1,30 2,15 3,10 5,6
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 30.
1+30=31 2+15=17 3+10=13 5+6=11
Calculate the sum for each pair.
a=3 b=10
The solution is the pair that gives sum 13.
\left(k^{2}+3k\right)+\left(10k+30\right)
Rewrite k^{2}+13k+30 as \left(k^{2}+3k\right)+\left(10k+30\right).
k\left(k+3\right)+10\left(k+3\right)
Factor out k in the first and 10 in the second group.
\left(k+3\right)\left(k+10\right)
Factor out common term k+3 by using distributive property.
k=-3 k=-10
To find equation solutions, solve k+3=0 and k+10=0.
k^{2}+13k+30=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.
k=\frac{-13±\sqrt{13^{2}-4\times 30}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 13 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-13±\sqrt{169-4\times 30}}{2}
Square 13.
k=\frac{-13±\sqrt{169-120}}{2}
Multiply -4 times 30.
k=\frac{-13±\sqrt{49}}{2}
Add 169 to -120.
k=\frac{-13±7}{2}
Take the square root of 49.
k=-\frac{6}{2}
Now solve the equation k=\frac{-13±7}{2} when ± is plus. Add -13 to 7.
k=-3
Divide -6 by 2.
k=-\frac{20}{2}
Now solve the equation k=\frac{-13±7}{2} when ± is minus. Subtract 7 from -13.
k=-10
Divide -20 by 2.
k=-3 k=-10
The equation is now solved.
k^{2}+13k+30=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.
k^{2}+13k+30-30=-30
Subtract 30 from both sides of the equation.
k^{2}+13k=-30
Subtracting 30 from itself leaves 0.
k^{2}+13k+\left(\frac{13}{2}\right)^{2}=-30+\left(\frac{13}{2}\right)^{2}
Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+13k+\frac{169}{4}=-30+\frac{169}{4}
Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.
k^{2}+13k+\frac{169}{4}=\frac{49}{4}
Add -30 to \frac{169}{4}.
\left(k+\frac{13}{2}\right)^{2}=\frac{49}{4}
Factor k^{2}+13k+\frac{169}{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(k+\frac{13}{2}\right)^{2}}=\sqrt{\frac{49}{4}}
Take the square root of both sides of the equation.
k+\frac{13}{2}=\frac{7}{2} k+\frac{13}{2}=-\frac{7}{2}
Simplify.
k=-3 k=-10
Subtract \frac{13}{2} from both sides of the equation.
x ^ 2 +13x +30 = 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 = -13 rs = 30
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{13}{2} - u s = -\frac{13}{2} + u
Two numbers r and s sum up to -13 exactly when the average of the two numbers is \frac{1}{2}*-13 = -\frac{13}{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{13}{2} - u) (-\frac{13}{2} + u) = 30
To solve for unknown quantity u, substitute these in the product equation rs = 30
\frac{169}{4} - u^2 = 30
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 30-\frac{169}{4} = -\frac{49}{4}
Simplify the expression by subtracting \frac{169}{4} on both sides
u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{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{13}{2} - \frac{7}{2} = -10 s = -\frac{13}{2} + \frac{7}{2} = -3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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