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x^{2}-x-\frac{9}{2}=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(-1\right)±\sqrt{1-4\left(-\frac{9}{2}\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -\frac{9}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1+18}}{2}
Multiply -4 times -\frac{9}{2}.
x=\frac{-\left(-1\right)±\sqrt{19}}{2}
Add 1 to 18.
x=\frac{1±\sqrt{19}}{2}
The opposite of -1 is 1.
x=\frac{\sqrt{19}+1}{2}
Now solve the equation x=\frac{1±\sqrt{19}}{2} when ± is plus. Add 1 to \sqrt{19}.
x=\frac{1-\sqrt{19}}{2}
Now solve the equation x=\frac{1±\sqrt{19}}{2} when ± is minus. Subtract \sqrt{19} from 1.
x=\frac{\sqrt{19}+1}{2} x=\frac{1-\sqrt{19}}{2}
The equation is now solved.
x^{2}-x-\frac{9}{2}=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.
x^{2}-x-\frac{9}{2}-\left(-\frac{9}{2}\right)=-\left(-\frac{9}{2}\right)
Add \frac{9}{2} to both sides of the equation.
x^{2}-x=-\left(-\frac{9}{2}\right)
Subtracting -\frac{9}{2} from itself leaves 0.
x^{2}-x=\frac{9}{2}
Subtract -\frac{9}{2} from 0.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{9}{2}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-x+\frac{1}{4}=\frac{9}{2}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-x+\frac{1}{4}=\frac{19}{4}
Add \frac{9}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{2}\right)^{2}=\frac{19}{4}
Factor x^{2}-x+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{\frac{19}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{19}}{2} x-\frac{1}{2}=-\frac{\sqrt{19}}{2}
Simplify.
x=\frac{\sqrt{19}+1}{2} x=\frac{1-\sqrt{19}}{2}
Add \frac{1}{2} to both sides of the equation.