Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 09:56, 1 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_3_2b.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (14:30, 26 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The first step when we solve the second-degree equation is to complete the square on the left-hand side
-	<center> [[Bild:2_3_2b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}
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		+	The equation can now be written as
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		+	{{Displayed math||<math>(y+1)^{2} = 16</math>}}
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		+	and has, after taking the square root, the solutions:
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		+	:*<math>y+1 = \sqrt{16} = 4\,\textrm{,}\ </math> which gives <math>y=-1+4=3\,\textrm{,}</math>
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		+	:*<math>y+1 = -\sqrt{16} = -4\,\textrm{,}\ </math> which gives <math>y=-1-4=-5\,\textrm{.}</math>
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		+	A quick check shows that <math>y=-5</math> and <math>y=3</math> satisfy the equation:
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		+	:*''y''&nbsp;=&nbsp;-5: <math>\ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}</math>
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		+	:*''y''&nbsp;=&nbsp;3: <math>\ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}</math>

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Current revision

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.