Solve for w
w=-1
w=4
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a+b=-3 ab=-4
To solve the equation, factor w^{2}-3w-4 using formula w^{2}+\left(a+b\right)w+ab=\left(w+a\right)\left(w+b\right). To find a and b, set up a system to be solved.
1,-4 2,-2
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 -4.
1-4=-3 2-2=0
Calculate the sum for each pair.
a=-4 b=1
The solution is the pair that gives sum -3.
\left(w-4\right)\left(w+1\right)
Rewrite factored expression \left(w+a\right)\left(w+b\right) using the obtained values.
w=4 w=-1
To find equation solutions, solve w-4=0 and w+1=0.
a+b=-3 ab=1\left(-4\right)=-4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as w^{2}+aw+bw-4. To find a and b, set up a system to be solved.
1,-4 2,-2
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 -4.
1-4=-3 2-2=0
Calculate the sum for each pair.
a=-4 b=1
The solution is the pair that gives sum -3.
\left(w^{2}-4w\right)+\left(w-4\right)
Rewrite w^{2}-3w-4 as \left(w^{2}-4w\right)+\left(w-4\right).
w\left(w-4\right)+w-4
Factor out w in w^{2}-4w.
\left(w-4\right)\left(w+1\right)
Factor out common term w-4 by using distributive property.
w=4 w=-1
To find equation solutions, solve w-4=0 and w+1=0.
w^{2}-3w-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.
w=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
w=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)}}{2}
Square -3.
w=\frac{-\left(-3\right)±\sqrt{9+16}}{2}
Multiply -4 times -4.
w=\frac{-\left(-3\right)±\sqrt{25}}{2}
Add 9 to 16.
w=\frac{-\left(-3\right)±5}{2}
Take the square root of 25.
w=\frac{3±5}{2}
The opposite of -3 is 3.
w=\frac{8}{2}
Now solve the equation w=\frac{3±5}{2} when ± is plus. Add 3 to 5.
w=4
Divide 8 by 2.
w=-\frac{2}{2}
Now solve the equation w=\frac{3±5}{2} when ± is minus. Subtract 5 from 3.
w=-1
Divide -2 by 2.
w=4 w=-1
The equation is now solved.
w^{2}-3w-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.
w^{2}-3w-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
w^{2}-3w=-\left(-4\right)
Subtracting -4 from itself leaves 0.
w^{2}-3w=4
Subtract -4 from 0.
w^{2}-3w+\left(-\frac{3}{2}\right)^{2}=4+\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
w^{2}-3w+\frac{9}{4}=4+\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
w^{2}-3w+\frac{9}{4}=\frac{25}{4}
Add 4 to \frac{9}{4}.
\left(w-\frac{3}{2}\right)^{2}=\frac{25}{4}
Factor w^{2}-3w+\frac{9}{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(w-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
w-\frac{3}{2}=\frac{5}{2} w-\frac{3}{2}=-\frac{5}{2}
Simplify.
w=4 w=-1
Add \frac{3}{2} to both sides of the equation.
x ^ 2 -3x -4 = 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 = 3 rs = -4
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{3}{2} - u s = \frac{3}{2} + u
Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{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{3}{2} - u) (\frac{3}{2} + u) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
\frac{9}{4} - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-\frac{9}{4} = -\frac{25}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{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{3}{2} - \frac{5}{2} = -1 s = \frac{3}{2} + \frac{5}{2} = 4
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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