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a+b=13 ab=15\times 2=30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15a^{2}+aa+ba+2. 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(15a^{2}+3a\right)+\left(10a+2\right)
Rewrite 15a^{2}+13a+2 as \left(15a^{2}+3a\right)+\left(10a+2\right).
3a\left(5a+1\right)+2\left(5a+1\right)
Factor out 3a in the first and 2 in the second group.
\left(5a+1\right)\left(3a+2\right)
Factor out common term 5a+1 by using distributive property.
a=-\frac{1}{5} a=-\frac{2}{3}
To find equation solutions, solve 5a+1=0 and 3a+2=0.
15a^{2}+13a+2=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.
a=\frac{-13±\sqrt{13^{2}-4\times 15\times 2}}{2\times 15}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, 13 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-13±\sqrt{169-4\times 15\times 2}}{2\times 15}
Square 13.
a=\frac{-13±\sqrt{169-60\times 2}}{2\times 15}
Multiply -4 times 15.
a=\frac{-13±\sqrt{169-120}}{2\times 15}
Multiply -60 times 2.
a=\frac{-13±\sqrt{49}}{2\times 15}
Add 169 to -120.
a=\frac{-13±7}{2\times 15}
Take the square root of 49.
a=\frac{-13±7}{30}
Multiply 2 times 15.
a=-\frac{6}{30}
Now solve the equation a=\frac{-13±7}{30} when ± is plus. Add -13 to 7.
a=-\frac{1}{5}
Reduce the fraction \frac{-6}{30} to lowest terms by extracting and canceling out 6.
a=-\frac{20}{30}
Now solve the equation a=\frac{-13±7}{30} when ± is minus. Subtract 7 from -13.
a=-\frac{2}{3}
Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.
a=-\frac{1}{5} a=-\frac{2}{3}
The equation is now solved.
15a^{2}+13a+2=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.
15a^{2}+13a+2-2=-2
Subtract 2 from both sides of the equation.
15a^{2}+13a=-2
Subtracting 2 from itself leaves 0.
\frac{15a^{2}+13a}{15}=-\frac{2}{15}
Divide both sides by 15.
a^{2}+\frac{13}{15}a=-\frac{2}{15}
Dividing by 15 undoes the multiplication by 15.
a^{2}+\frac{13}{15}a+\left(\frac{13}{30}\right)^{2}=-\frac{2}{15}+\left(\frac{13}{30}\right)^{2}
Divide \frac{13}{15}, the coefficient of the x term, by 2 to get \frac{13}{30}. Then add the square of \frac{13}{30} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}+\frac{13}{15}a+\frac{169}{900}=-\frac{2}{15}+\frac{169}{900}
Square \frac{13}{30} by squaring both the numerator and the denominator of the fraction.
a^{2}+\frac{13}{15}a+\frac{169}{900}=\frac{49}{900}
Add -\frac{2}{15} to \frac{169}{900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a+\frac{13}{30}\right)^{2}=\frac{49}{900}
Factor a^{2}+\frac{13}{15}a+\frac{169}{900}. 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(a+\frac{13}{30}\right)^{2}}=\sqrt{\frac{49}{900}}
Take the square root of both sides of the equation.
a+\frac{13}{30}=\frac{7}{30} a+\frac{13}{30}=-\frac{7}{30}
Simplify.
a=-\frac{1}{5} a=-\frac{2}{3}
Subtract \frac{13}{30} from both sides of the equation.
x ^ 2 +\frac{13}{15}x +\frac{2}{15} = 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 15
r + s = -\frac{13}{15} rs = \frac{2}{15}
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}{30} - u s = -\frac{13}{30} + u
Two numbers r and s sum up to -\frac{13}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{15} = -\frac{13}{30}. 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}{30} - u) (-\frac{13}{30} + u) = \frac{2}{15}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{15}
\frac{169}{900} - u^2 = \frac{2}{15}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{15}-\frac{169}{900} = -\frac{49}{900}
Simplify the expression by subtracting \frac{169}{900} on both sides
u^2 = \frac{49}{900} u = \pm\sqrt{\frac{49}{900}} = \pm \frac{7}{30}
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}{30} - \frac{7}{30} = -0.667 s = -\frac{13}{30} + \frac{7}{30} = -0.200
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.