x 2 1 2x 1

x^{2}-\frac{1}{2}x-1=0

All equations of the khuông 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(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\left(-1\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{1}{2} for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-4\left(-1\right)}}{2}

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}+4}}{2}

Multiply -4 times -1.

x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{17}{4}}}{2}

Add \frac{1}{4} to lớn 4.

x=\frac{-\left(-\frac{1}{2}\right)±\frac{\sqrt{17}}{2}}{2}

Take the square root of \frac{17}{4}.

x=\frac{\frac{1}{2}±\frac{\sqrt{17}}{2}}{2}

The opposite of -\frac{1}{2} is \frac{1}{2}.

x=\frac{\sqrt{17}+1}{2\times 2}

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{17}}{2}}{2} when ± is plus. Add \frac{1}{2} to lớn \frac{\sqrt{17}}{2}.

x=\frac{\sqrt{17}+1}{4}

Divide \frac{1+\sqrt{17}}{2} by 2.

x=\frac{1-\sqrt{17}}{2\times 2}

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{17}}{2}}{2} when ± is minus. Subtract \frac{\sqrt{17}}{2} from \frac{1}{2}.

x=\frac{1-\sqrt{17}}{4}

Divide \frac{1-\sqrt{17}}{2} by 2.

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x=\frac{\sqrt{17}+1}{4} x=\frac{1-\sqrt{17}}{4}

The equation is now solved.

x^{2}-\frac{1}{2}x-1=0

Quadratic equations such as this one can be solved by completing the square. In order to lớn complete the square, the equation must first be in the khuông x^{2}+bx=c.

x^{2}-\frac{1}{2}x-1-\left(-1\right)=-\left(-1\right)

Add 1 to lớn both sides of the equation.

x^{2}-\frac{1}{2}x=-\left(-1\right)

Subtracting -1 from itself leaves 0.

x^{2}-\frac{1}{2}x=1

Subtract -1 from 0.

x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=1+\left(-\frac{1}{4}\right)^{2}

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to lớn get -\frac{1}{4}. Then add the square of -\frac{1}{4} to lớn both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-\frac{1}{2}x+\frac{1}{16}=1+\frac{1}{16}

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{17}{16}

Add 1 to lớn \frac{1}{16}.

\left(x-\frac{1}{4}\right)^{2}=\frac{17}{16}

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}{4}\right)^{2}}=\sqrt{\frac{17}{16}}

Take the square root of both sides of the equation.

x-\frac{1}{4}=\frac{\sqrt{17}}{4} x-\frac{1}{4}=-\frac{\sqrt{17}}{4}

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Simplify.

x=\frac{\sqrt{17}+1}{4} x=\frac{1-\sqrt{17}}{4}

Add \frac{1}{4} to lớn both sides of the equation.