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x^{2}-3x-4=0
Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.
a+b=-3 ab=-4
To solve the equation, factor x^{2}-3x-4 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+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(x-4\right)\left(x+1\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=4 x=-1
To find equation solutions, solve x-4=0 and x+1=0.
x=-1
Variable x cannot be equal to 4.
x^{2}-3x-4=0
Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.
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 x^{2}+ax+bx-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(x^{2}-4x\right)+\left(x-4\right)
Rewrite x^{2}-3x-4 as \left(x^{2}-4x\right)+\left(x-4\right).
x\left(x-4\right)+x-4
Factor out x in x^{2}-4x.
\left(x-4\right)\left(x+1\right)
Factor out common term x-4 by using distributive property.
x=4 x=-1
To find equation solutions, solve x-4=0 and x+1=0.
x=-1
Variable x cannot be equal to 4.
x^{2}-3x-4=0
Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.
x=\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}.
x=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)}}{2}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9+16}}{2}
Multiply -4 times -4.
x=\frac{-\left(-3\right)±\sqrt{25}}{2}
Add 9 to 16.
x=\frac{-\left(-3\right)±5}{2}
Take the square root of 25.
x=\frac{3±5}{2}
The opposite of -3 is 3.
x=\frac{8}{2}
Now solve the equation x=\frac{3±5}{2} when ± is plus. Add 3 to 5.
x=4
Divide 8 by 2.
x=-\frac{2}{2}
Now solve the equation x=\frac{3±5}{2} when ± is minus. Subtract 5 from 3.
x=-1
Divide -2 by 2.
x=4 x=-1
The equation is now solved.
x=-1
Variable x cannot be equal to 4.
x^{2}-3x-4=0
Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.
x^{2}-3x=4
Add 4 to both sides. Anything plus zero gives itself.
x^{2}-3x+\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.
x^{2}-3x+\frac{9}{4}=4+\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-3x+\frac{9}{4}=\frac{25}{4}
Add 4 to \frac{9}{4}.
\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}
Factor x^{2}-3x+\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(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
x-\frac{3}{2}=\frac{5}{2} x-\frac{3}{2}=-\frac{5}{2}
Simplify.
x=4 x=-1
Add \frac{3}{2} to both sides of the equation.
x=-1
Variable x cannot be equal to 4.