Solve for y
y=-\frac{1}{3}\approx -0.333333333
y=1
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a+b=-2 ab=3\left(-1\right)=-3
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3y^{2}+ay+by-1. To find a and b, set up a system to be solved.
a=-3 b=1
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
\left(3y^{2}-3y\right)+\left(y-1\right)
Rewrite 3y^{2}-2y-1 as \left(3y^{2}-3y\right)+\left(y-1\right).
3y\left(y-1\right)+y-1
Factor out 3y in 3y^{2}-3y.
\left(y-1\right)\left(3y+1\right)
Factor out common term y-1 by using distributive property.
y=1 y=-\frac{1}{3}
To find equation solutions, solve y-1=0 and 3y+1=0.
3y^{2}-2y-1=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.
y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-1\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-1\right)}}{2\times 3}
Square -2.
y=\frac{-\left(-2\right)±\sqrt{4-12\left(-1\right)}}{2\times 3}
Multiply -4 times 3.
y=\frac{-\left(-2\right)±\sqrt{4+12}}{2\times 3}
Multiply -12 times -1.
y=\frac{-\left(-2\right)±\sqrt{16}}{2\times 3}
Add 4 to 12.
y=\frac{-\left(-2\right)±4}{2\times 3}
Take the square root of 16.
y=\frac{2±4}{2\times 3}
The opposite of -2 is 2.
y=\frac{2±4}{6}
Multiply 2 times 3.
y=\frac{6}{6}
Now solve the equation y=\frac{2±4}{6} when ± is plus. Add 2 to 4.
y=1
Divide 6 by 6.
y=-\frac{2}{6}
Now solve the equation y=\frac{2±4}{6} when ± is minus. Subtract 4 from 2.
y=-\frac{1}{3}
Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.
y=1 y=-\frac{1}{3}
The equation is now solved.
3y^{2}-2y-1=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.
3y^{2}-2y-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
3y^{2}-2y=-\left(-1\right)
Subtracting -1 from itself leaves 0.
3y^{2}-2y=1
Subtract -1 from 0.
\frac{3y^{2}-2y}{3}=\frac{1}{3}
Divide both sides by 3.
y^{2}-\frac{2}{3}y=\frac{1}{3}
Dividing by 3 undoes the multiplication by 3.
y^{2}-\frac{2}{3}y+\left(-\frac{1}{3}\right)^{2}=\frac{1}{3}+\left(-\frac{1}{3}\right)^{2}
Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-\frac{2}{3}y+\frac{1}{9}=\frac{1}{3}+\frac{1}{9}
Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.
y^{2}-\frac{2}{3}y+\frac{1}{9}=\frac{4}{9}
Add \frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y-\frac{1}{3}\right)^{2}=\frac{4}{9}
Factor y^{2}-\frac{2}{3}y+\frac{1}{9}. 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(y-\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}
Take the square root of both sides of the equation.
y-\frac{1}{3}=\frac{2}{3} y-\frac{1}{3}=-\frac{2}{3}
Simplify.
y=1 y=-\frac{1}{3}
Add \frac{1}{3} to both sides of the equation.
x ^ 2 -\frac{2}{3}x -\frac{1}{3} = 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 3
r + s = \frac{2}{3} rs = -\frac{1}{3}
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{1}{3} - u s = \frac{1}{3} + u
Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. 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{1}{3} - u) (\frac{1}{3} + u) = -\frac{1}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}
\frac{1}{9} - u^2 = -\frac{1}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{3}-\frac{1}{9} = -\frac{4}{9}
Simplify the expression by subtracting \frac{1}{9} on both sides
u^2 = \frac{4}{9} u = \pm\sqrt{\frac{4}{9}} = \pm \frac{2}{3}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{3} - \frac{2}{3} = -0.333 s = \frac{1}{3} + \frac{2}{3} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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Linear equation
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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
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Limits
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