Solve for x
x=\frac{2}{5}=0.4
x=\frac{2}{3}\approx 0.666666667
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a+b=-16 ab=15\times 4=60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx+4. To find a and b, set up a system to be solved.
-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10
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 60.
-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16
Calculate the sum for each pair.
a=-10 b=-6
The solution is the pair that gives sum -16.
\left(15x^{2}-10x\right)+\left(-6x+4\right)
Rewrite 15x^{2}-16x+4 as \left(15x^{2}-10x\right)+\left(-6x+4\right).
5x\left(3x-2\right)-2\left(3x-2\right)
Factor out 5x in the first and -2 in the second group.
\left(3x-2\right)\left(5x-2\right)
Factor out common term 3x-2 by using distributive property.
x=\frac{2}{3} x=\frac{2}{5}
To find equation solutions, solve 3x-2=0 and 5x-2=0.
15x^{2}-16x+4=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(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 15\times 4}}{2\times 15}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -16 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-16\right)±\sqrt{256-4\times 15\times 4}}{2\times 15}
Square -16.
x=\frac{-\left(-16\right)±\sqrt{256-60\times 4}}{2\times 15}
Multiply -4 times 15.
x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 15}
Multiply -60 times 4.
x=\frac{-\left(-16\right)±\sqrt{16}}{2\times 15}
Add 256 to -240.
x=\frac{-\left(-16\right)±4}{2\times 15}
Take the square root of 16.
x=\frac{16±4}{2\times 15}
The opposite of -16 is 16.
x=\frac{16±4}{30}
Multiply 2 times 15.
x=\frac{20}{30}
Now solve the equation x=\frac{16±4}{30} when ± is plus. Add 16 to 4.
x=\frac{2}{3}
Reduce the fraction \frac{20}{30} to lowest terms by extracting and canceling out 10.
x=\frac{12}{30}
Now solve the equation x=\frac{16±4}{30} when ± is minus. Subtract 4 from 16.
x=\frac{2}{5}
Reduce the fraction \frac{12}{30} to lowest terms by extracting and canceling out 6.
x=\frac{2}{3} x=\frac{2}{5}
The equation is now solved.
15x^{2}-16x+4=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.
15x^{2}-16x+4-4=-4
Subtract 4 from both sides of the equation.
15x^{2}-16x=-4
Subtracting 4 from itself leaves 0.
\frac{15x^{2}-16x}{15}=-\frac{4}{15}
Divide both sides by 15.
x^{2}-\frac{16}{15}x=-\frac{4}{15}
Dividing by 15 undoes the multiplication by 15.
x^{2}-\frac{16}{15}x+\left(-\frac{8}{15}\right)^{2}=-\frac{4}{15}+\left(-\frac{8}{15}\right)^{2}
Divide -\frac{16}{15}, the coefficient of the x term, by 2 to get -\frac{8}{15}. Then add the square of -\frac{8}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{16}{15}x+\frac{64}{225}=-\frac{4}{15}+\frac{64}{225}
Square -\frac{8}{15} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{16}{15}x+\frac{64}{225}=\frac{4}{225}
Add -\frac{4}{15} to \frac{64}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{8}{15}\right)^{2}=\frac{4}{225}
Factor x^{2}-\frac{16}{15}x+\frac{64}{225}. 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{8}{15}\right)^{2}}=\sqrt{\frac{4}{225}}
Take the square root of both sides of the equation.
x-\frac{8}{15}=\frac{2}{15} x-\frac{8}{15}=-\frac{2}{15}
Simplify.
x=\frac{2}{3} x=\frac{2}{5}
Add \frac{8}{15} to both sides of the equation.
x ^ 2 -\frac{16}{15}x +\frac{4}{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{16}{15} rs = \frac{4}{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{8}{15} - u s = \frac{8}{15} + u
Two numbers r and s sum up to \frac{16}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{15} = \frac{8}{15}. 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{8}{15} - u) (\frac{8}{15} + u) = \frac{4}{15}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{15}
\frac{64}{225} - u^2 = \frac{4}{15}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{4}{15}-\frac{64}{225} = -\frac{4}{225}
Simplify the expression by subtracting \frac{64}{225} on both sides
u^2 = \frac{4}{225} u = \pm\sqrt{\frac{4}{225}} = \pm \frac{2}{15}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{8}{15} - \frac{2}{15} = 0.400 s = \frac{8}{15} + \frac{2}{15} = 0.667
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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