Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 2.3:2b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (14:30, 26 September 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
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
 +
{{Displayed math||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}
-
<math>y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.</math>
+
The equation can now be written as
-
The equation can now be written as
+
{{Displayed math||<math>(y+1)^{2} = 16</math>}}
 +
and has, after taking the square root, the solutions:
-
<math>\left( y+1 \right)^{2}=16</math>
+
:*<math>y+1 = \sqrt{16} = 4\,\textrm{,}\ </math> which gives <math>y=-1+4=3\,\textrm{,}</math>
 +
:*<math>y+1 = -\sqrt{16} = -4\,\textrm{,}\ </math> which gives <math>y=-1-4=-5\,\textrm{.}</math>
-
and has, after taking the square root, the solutions
 
 +
A quick check shows that <math>y=-5</math> and <math>y=3</math> satisfy the equation:
-
<math>y+1=\sqrt{16}=4</math>
+
:*''y''&nbsp;=&nbsp;-5: <math>\ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}</math>
-
which gives
+
-
<math>y=-1+4=3</math>
+
-
 
+
:*''y''&nbsp;=&nbsp;3: <math>\ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}</math>
-
<math>y+1=-\sqrt{16}=-4</math>
+
-
which gives
+
-
<math>y=-1-4=-5</math>
+
-
 
+
-
 
+
-
A quick check shows that
+
-
<math>y=-\text{5 }</math>
+
-
and
+
-
<math>y=\text{3 }</math>
+
-
satisfy the equation:
+
-
 
+
-
 
+
-
<math>y=-\text{5 }</math>: LHS=
+
-
<math>\left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0</math>
+
-
= RHS
+
-
 
+
-
<math>y=\text{3 }</math>: LHS=
+
-
<math>3^{2}+2\centerdot 3-15=9+6-15=0</math>
+
-
= RHS
+

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.
Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.3:2b"