Solution 2.3:2a

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m (Lösning 2.3:2a moved to Solution 2.3:2a: Robot: moved page)
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We solve the second order equation by combining together the x2- and
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<center> [[Image:2_3_2a.gif]] </center>
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<math>x</math>
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-terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.
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By completing the square, the left-hand side becomes
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<math>\underline{x^{2}-4x}+3=\underline{\left( x-2 \right)^{2}-2^{2}}+3=\left( x-2 \right)^{2}-1</math>
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where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as
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<math>\left( x-2 \right)^{2}-1=0</math>
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which we solve by moving the "
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<math>1</math>
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" on the right-hand side and taking the square root. This gives the solutions
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<math>x-2=\sqrt{1}=1</math>
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i.e.
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<math>x=2+1=3</math>
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<math>x-2=-\sqrt{1}=-1</math>
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i.e.
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<math>x=2-1=1</math>
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Because it is easy to make a mistake, we check the answer by substituting
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<math>x=1</math>
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and
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<math>x=3</math>
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into the original equation.:
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<math>x=\text{1}</math>: LHS=
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<math>1^{2}-4\centerdot 1+3=1-4+3=0</math>
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= RHS
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<math>x=3</math>: LHS=
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<math>3^{2}-4\centerdot 3+3=9-12+3=0</math>
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= RHS

Revision as of 13:21, 20 September 2008

We solve the second order equation by combining together the x2- and \displaystyle x -terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.

By completing the square, the left-hand side becomes


\displaystyle \underline{x^{2}-4x}+3=\underline{\left( x-2 \right)^{2}-2^{2}}+3=\left( x-2 \right)^{2}-1


where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as


\displaystyle \left( x-2 \right)^{2}-1=0


which we solve by moving the " \displaystyle 1 " on the right-hand side and taking the square root. This gives the solutions


\displaystyle x-2=\sqrt{1}=1 i.e. \displaystyle x=2+1=3


\displaystyle x-2=-\sqrt{1}=-1 i.e. \displaystyle x=2-1=1


Because it is easy to make a mistake, we check the answer by substituting \displaystyle x=1 and \displaystyle x=3 into the original equation.:


\displaystyle x=\text{1}: LHS= \displaystyle 1^{2}-4\centerdot 1+3=1-4+3=0 = RHS

\displaystyle x=3: LHS= \displaystyle 3^{2}-4\centerdot 3+3=9-12+3=0 = RHS

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