- COMMON ALGEBRA ERRORS–
968.Since taking the square root of both sides of the
inequality x^2 4 yields the statement x2.
Since both statements must be satisfied simul-
taneously, the solution set is (–∞, –2].
a.You must move all terms to one side of the
inequality, factor (if possible), determine the
values that make the factored expression
equal to zero, and construct a sign chart to
solve such an inequality. The correct solution
set should be [–2,2].
b.When taking the square root of both sides
of an equation, you use only the principal
root. As such, the correct statement should
be x 2, so that the solution set is (–∞,2].
c.There is no error.
969.(x– y)^2 = x^2 – y^2
a.The left side must be expanded by FOILing.
The correct statement should be (x– y)^2 =
x^2 – 2xy +y^2.
b.The –1 must be squared. The correct state-
ment should be (x– y)^2 = x^2 + y^2.
c.There is no error.
970.The solution of the equation x= –2 is x= 4,
as seen by squaring both sides of the equation.
a.The correct solution is x= –4 because when
you square both sides of the equation, you
do not square the –1.
b.This equation has no real solutions because
the output of an even–indexed radical must
be nonnegative.
c.There is no error.
971.The solution set of the inequality |x+ 2| 5
is (–∞, –7)∪(3,∞).
a.The interval (–∞, –7) should be deleted
because an absolute value inequality cannot
have negative solutions.
b.You must include the values that make the
left side equal to 5. As such, the solution set
should be (–∞, –7]∪[3,∞)
c.There is no error.
972.x^2 + 25 = (x– 5)(x+ 5)
a.The correct factorization of the left side is
x^2 + 25 = x^2 + 52 = (x+ 5)^2.
b.The left side is not a difference of squares. It
cannot be factored further.
c.There is no error.973. = = ^15
a.Cancelling the terms x–1 and y–1 leaves 0
each time, not 1. So, the correct statement
should be = ^24 = ^12 .
b.You cannot cancel terms of a sum in the
numerator and denominator. You can
only cancel factors common to both. The
complex fraction must first be simplified
before any cancelation can occur. The
correct statement is:= = = =2yx – yx = y2y+– 4 xxc.There is no error.974.ln(ex+ e2y) = ln(ex) + ln(e2y) = x+ 2y
a.The first equality is incorrect because the
natural logarithm of a sum is the product of
the natural logarithms. So, the statement
should be ln(ex+ e2y) = ln(ex)ln(e2y) = 2xy.
b.The first equality is incorrect because the
natural logarithm of a sum is not the sum of
the natural logarithms. In fact, the expression
on the extreme left side of the string of
equalities cannot be simplified further.
c.There is no error.xy^
y+4x^2 yx–yx
y+xy4x^2 xyy– xxy
xyy+ ^4 xxy^2 x– ^1 y
1
x+ ^4 y^2 x–1– y–1
x–1+ 4y–1^2 x^ –1– y^ –1
x –1+4y –12 – 1
1 + 4
^2 x^ –1– y^ –1
x –1+ 4y –1