Question

question 2
2.1 given the quadratic equation 2x² - 2x + 1 = 0.
2.1.1 solve the equation. (4)
2.1.2 show the roots of the equation on an argand diagram. (2)
2.1.3 rewrite the roots in polar form with the angles in radians. (4)

Explanation:

Step1: Recall quadratic - formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). In the equation \(2x^{2}-2x + 1=0\), \(a = 2\), \(b=-2\), and \(c = 1\). First, calculate the discriminant \(\Delta=b^{2}-4ac\).

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Step2: Find the roots

Substitute into the quadratic - formula:

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Step3: Plot on Argand diagram

The roots \(x_1=\frac{1}{2}+\frac{1}{2}i\) and \(x_2=\frac{1}{2}-\frac{1}{2}i\) are complex numbers. In the Argand diagram, for a complex number \(z=x + yi\), \(x\) is the real - part and \(y\) is the imaginary - part. For \(z_1=\frac{1}{2}+\frac{1}{2}i\), the point is \((\frac{1}{2},\frac{1}{2})\) and for \(z_2=\frac{1}{2}-\frac{1}{2}i\), the point is \((\frac{1}{2},-\frac{1}{2})\).

Step4: Convert to polar form

For a complex number \(z=x + yi\), \(r=\sqrt{x^{2}+y^{2}}\) and \(\theta=\tan^{- 1}(\frac{y}{x})\).
For \(z_1=\frac{1}{2}+\frac{1}{2}i\):

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So, \(z_1=\frac{\sqrt{2}}{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})\).
For \(z_2=\frac{1}{2}-\frac{1}{2}i\):

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So, \(z_2=\frac{\sqrt{2}}{2}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))\).

Answer:

2.1.1 \(x=\frac{1}{2}\pm\frac{1}{2}i\)
2.1.2 Plot the points \((\frac{1}{2},\frac{1}{2})\) and \((\frac{1}{2},-\frac{1}{2})\) on the Argand diagram.
2.1.3 \(z_1=\frac{\sqrt{2}}{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})\), \(z_2=\frac{\sqrt{2}}{2}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))\)