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Lösung 2.3:2b

Aus Online Mathematik Brückenkurs 1

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The first step when we solve the second-degree equation is to complete the square on the left-hand side
The first step when we solve the second-degree equation is to complete the square on the left-hand side
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{{Displayed math||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}
The equation can now be written as
The equation can now be written as
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{{Displayed math||<math>(y+1)^{2} = 16</math>}}
+
{{Abgesetzte Formel||<math>(y+1)^{2} = 16</math>}}
and has, after taking the square root, the solutions:
and has, after taking the square root, the solutions:

Version vom 08:31, 22. Okt. 2008

The first step when we solve the second-degree equation is to complete the square on the left-hand side

y2+2y−15=(y+1)2−12−15=(y+1)2−16.

The equation can now be written as

(y+1)2=16

and has, after taking the square root, the solutions:

  • y+1=16=4,   which gives y=−1+4=3,
  • y+1=−16=−4,   which gives y=−1−4=−5.


A quick check shows that y=−5 and y=3 satisfy the equation:

  • y = -5:  LHS=(−5)2+2(−5)−15=25−10−15=0=RHS,
  • y = 3:  LHS=32+23−15=9+6−15=0=RHS.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:2b“