Solve for y
y=-3
y=8
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a+b=-5 ab=-24
To solve the equation, factor y^{2}-5y-24 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.
1,-24 2,-12 3,-8 4,-6
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. List all such integer pairs that give product -24.
1-24=-23 2-12=-10 3-8=-5 4-6=-2
Calculate the sum for each pair.
a=-8 b=3
The solution is the pair that gives sum -5.
\left(y-8\right)\left(y+3\right)
Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.
y=8 y=-3
To find equation solutions, solve y-8=0 and y+3=0.
a+b=-5 ab=1\left(-24\right)=-24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-24. To find a and b, set up a system to be solved.
1,-24 2,-12 3,-8 4,-6
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. List all such integer pairs that give product -24.
1-24=-23 2-12=-10 3-8=-5 4-6=-2
Calculate the sum for each pair.
a=-8 b=3
The solution is the pair that gives sum -5.
\left(y^{2}-8y\right)+\left(3y-24\right)
Rewrite y^{2}-5y-24 as \left(y^{2}-8y\right)+\left(3y-24\right).
y\left(y-8\right)+3\left(y-8\right)
Factor out y in the first and 3 in the second group.
\left(y-8\right)\left(y+3\right)
Factor out common term y-8 by using distributive property.
y=8 y=-3
To find equation solutions, solve y-8=0 and y+3=0.
y^{2}-5y-24=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-24\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-5\right)±\sqrt{25-4\left(-24\right)}}{2}
Square -5.
y=\frac{-\left(-5\right)±\sqrt{25+96}}{2}
Multiply -4 times -24.
y=\frac{-\left(-5\right)±\sqrt{121}}{2}
Add 25 to 96.
y=\frac{-\left(-5\right)±11}{2}
Take the square root of 121.
y=\frac{5±11}{2}
The opposite of -5 is 5.
y=\frac{16}{2}
Now solve the equation y=\frac{5±11}{2} when ± is plus. Add 5 to 11.
y=8
Divide 16 by 2.
y=-\frac{6}{2}
Now solve the equation y=\frac{5±11}{2} when ± is minus. Subtract 11 from 5.
y=-3
Divide -6 by 2.
y=8 y=-3
The equation is now solved.
y^{2}-5y-24=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.
y^{2}-5y-24-\left(-24\right)=-\left(-24\right)
Add 24 to both sides of the equation.
y^{2}-5y=-\left(-24\right)
Subtracting -24 from itself leaves 0.
y^{2}-5y=24
Subtract -24 from 0.
y^{2}-5y+\left(-\frac{5}{2}\right)^{2}=24+\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-5y+\frac{25}{4}=24+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
y^{2}-5y+\frac{25}{4}=\frac{121}{4}
Add 24 to \frac{25}{4}.
\left(y-\frac{5}{2}\right)^{2}=\frac{121}{4}
Factor y^{2}-5y+\frac{25}{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(y-\frac{5}{2}\right)^{2}}=\sqrt{\frac{121}{4}}
Take the square root of both sides of the equation.
y-\frac{5}{2}=\frac{11}{2} y-\frac{5}{2}=-\frac{11}{2}
Simplify.
y=8 y=-3
Add \frac{5}{2} to both sides of the equation.
x ^ 2 -5x -24 = 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 = 5 rs = -24
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{5}{2} - u s = \frac{5}{2} + u
Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{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{5}{2} - u) (\frac{5}{2} + u) = -24
To solve for unknown quantity u, substitute these in the product equation rs = -24
\frac{25}{4} - u^2 = -24
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -24-\frac{25}{4} = -\frac{121}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{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{5}{2} - \frac{11}{2} = -3 s = \frac{5}{2} + \frac{11}{2} = 8
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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