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Lösung 3.3:4b

Aus Online Mathematik Brückenkurs 2

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If we complete the square of the left-hand side, we get
If we complete the square of the left-hand side, we get
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(z-2)^2-2^2+5&=0,\\[5pt]
(z-2)^2-2^2+5&=0,\\[5pt]
(z-2)^2+1&=0.
(z-2)^2+1&=0.

Version vom 13:13, 10. Mär. 2009

Typically, one solves a second-degree by completing the square, followed by taking the square root.

If we complete the square of the left-hand side, we get

(z−2)2−22+5(z−2)2+1=0=0

Taking the square root then gives that the equation has roots z−2=i, i.e. z=2+i and z=2−i.

If we want to be sure that we have found the correct solutions, we can substitute each solution into the equation and see whether the equation is satisfied.

z=2+i:z2−4z+5z=2−i:z2−4z+5=(2+i)2−4(2+i)+5=22+4i+i2−8−4i+5=4+4i−1−8−4i+5=0=(2−i)2−4(2−i)+5=22−4i+i2−8+4i+5=4−4i−1−8+4i+5=0.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:4b“