Solution 3.2:2
From Förberedande kurs i matematik 1
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<math>2x+7=x^{2}+4x+4</math> | <math>2x+7=x^{2}+4x+4</math> | ||
Revision as of 10:30, 25 September 2008
The first thing we do is to square both sides of the equation
\displaystyle 2x+7=\left( x+2 \right)^{2}
to get an equation without a root sign. It is possible that we thereby introduce so-called false roots (solutions to the new equation which are not solutions to the old equation), so we need to test the solutions in the original root equation before we answer.
If we expand the right-hand side in the squared equation, we get
(*)
\displaystyle 2x+7=x^{2}+4x+4
which we also can write as
\displaystyle x^{2}+2x-3=0
Completing the square of the left-hand side gives
\displaystyle x^{2}+2x-3=\left( x+1 \right)^{2}-1^{2}-3=\left( x+1 \right)^{2}-4.
The equation then becomes
\displaystyle \left( x+1 \right)^{2}=4
which has solutions
\displaystyle x=-1+\sqrt{4}=-1+2=1
\displaystyle x=-1-\sqrt{4}=-1-2=-3
A quick check shows also that
\displaystyle x=-\text{3 }
and
\displaystyle x=\text{1 }
are solutions to the squared equation (*)
\displaystyle x=-\text{3}: LHS
\displaystyle =2\centerdot \left( -3 \right)+7=-6+7=1
and
RHS
\displaystyle =\left( -3+2 \right)^{2}=1
\displaystyle x=\text{1}: LHS
\displaystyle =2\centerdot 1+7=2+7=9
and
RHS
\displaystyle =\left( 1+2 \right)^{2}=9
When we test solution in the root equation, we get that
\displaystyle x=-\text{3}: LHS
\displaystyle =\sqrt{2\centerdot \left( -3 \right)+7}=\sqrt{-6+7}=1
RHS \displaystyle =-3+2=-1
\displaystyle x=\text{1}: LHS
\displaystyle =\sqrt{2\centerdot 1+7}=\sqrt{2+7}=3
RHS \displaystyle =1+2=3
and therefore
\displaystyle x=\text{1}
is the only solution to the root equation (
\displaystyle x=-\text{3}
is a false root).
NOTE: The check we carry out when substituting the solutions into equation (*) is not strictly speaking necessary, but more for seeing that we haven't calculated incorrectly. On the other hand, testing in the root equation is necessary.
