Lösung 3.2:2

Aus Online Mathematik Brückenkurs 1

The first thing we do is to square both sides of the equation

to obtain an equation without a root sign. It is possible that we thereby introduce so-called spurious 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

which we also can write as

Completing the square of the left-hand side gives

The equation then becomes

which has solutions

• x=−1+4=−1+2=1 

• x=−1−4=−1−2=−3 

A quick check shows also that x=−3 and x=1 are solutions to the squared equation (*):

x = -3:	LHS=2(−3)+7=−6+7=1 and
	RHS=(−3+2)2=1
x = 1:	LHS=21+7=2+7=9 and
	RHS=(1+2)2=9

When we test the solutions in the root equation, we get that

x = -3:	LHS=2(−3)+7=−6+7=1=1  and
	RHS=−3+2=−1
x = 1:	LHS=21+7=2+7=9=3
	RHS=1+2=3

and therefore x=1 is the only solution to the root equation (x=−3 is a spurious 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.