SME
Grade 11 Math · Problem type
Extraneous solutions
Use this for “Solve √(x + 7) = x + 1” and reject the extraneous solution.
Chart
Zeros: x = −3, x = 2
Left side at x₂: 2
Right side at x₂: −2
This is an example. Type your own values to start a new problem.
x
+
=
x
+
b
x²
+
B
x
+
(
C
)
=
0
Square both sides: x + a = (x + b)², which is the second line. The graph shows its zeros.
x₁
Larger candidate
worked out (not typed)
x₂
Smaller candidate
worked out (not typed)
L
Left side at x₂
worked out (not typed)
R
Right side at x₂
worked out (not typed)
Formulas
B
=
2
×
b
−
1
1 = 2 × 1 − 1
C
=
b
2
−
a
−6
=
1
2
−
7
x₁
=
−
B
+
√
B
2
− 4 ×
C
2
2
=
(−1
+
√(
1
2
−
4
×
(−6)))
÷
2
x₂
=
−
B
−
√
B
2
− 4 ×
C
2
−3
=
(−1
−
√(
1
2
−
4
×
(−6)))
÷
2
L
=
√
x₂
+
a
2
=
√
−3 + 7
R
=
x₂
+
b
−2 = −3 + 1
Assumptions
•
Squaring both sides gives x + a = (x + b)², a quadratic: the graph shows its two zeros, the candidates.
•
Squaring can add a false solution: one where x + b is negative, since a root never is.
•
Check each candidate in the first equation; keep only the ones that work.
Step-by-step
WE KNOW
a = 7, b = 1
FIND
x coefficient after squaring, 2b − 1 (B), constant after squaring, b² − a (C), larger candidate (x₁), smaller candidate (x₂), left side at x₂ (L), right side at x₂ (R)
Step 1 · Find x coefficient after squaring, 2b − 1 (B)
Use B = 2 × b − 1
Square
both
sides:
x
+
a
=
x
2
+
2bx
+
b
2
,
so
x
2
+
(
2
b
−
1)x
+
(
b
2
−
a
)
=
0.
B
=
2
×
1
−
1
B
=
1
Step 2 · Find constant after squaring, b² − a (C)
Use C = b² − a
The constant of the squared equation.
C
=
1
2
−
7
C
=
1
−
7
C
=
−6
Step 3 · Find larger candidate (x₁)
Use x₁ = (−B + √(B² − 4 × C)) ÷ 2
x
2
+
Bx
+
C
=
0
by
the
quadratic
formula;
here
x
+
b
>
0,
so
it
always
works.
x₁
=
(−1
+
√(
1
2
−
4
×
(−6)))
÷
2
x₁
=
−1 +
√
25
2
x₁
=
−1 + 5
2
x₁
=
2
Step 4 · Find smaller candidate (x₂)
Use x₂ = (−B − √(B² − 4 × C)) ÷ 2
The other candidate, with the minus sign.
x₂
=
(−1
−
√(
1
2
−
4
×
(−6)))
÷
2
x₂
=
−1 −
√
25
2
x₂
=
−1 − 5
2
x₂
=
−3
Step 5 · Find left side at x₂ (L)
Use L = √(x₂ + a)
Check
x₂
in
the
left
side
of
the
first
equation.
L
=
√
−3 + 7
L
=
√
4
L
=
2
Step 6 · Find right side at x₂ (R)
Use R = x₂ + b
The
right
side
is
negative
but
a
root
is
not:
x₂
is
extraneous,
so
reject
it.
R
=
−3
+
1
R
=
−2
Check
Put the numbers back in. Both sides should match.
1 = 2 × 1 − 1 ✓
−6
=
1
2
−
7
✓
2
=
(−1
+
√(
1
2
−
4
×
(−6)))
÷
2
✓
−3
=
(−1
−
√(
1
2
−
4
×
(−6)))
÷
2
✓
2
=
√
−3 + 7
✓
−2 = −3 + 1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Radical functions and equations
Main lesson
Graph a square root function
Use this for “Graph y = 2√(x + 3) + 1: where it starts, which way it goes, and y at x = 6.”
Graph a cube root function
Use this for “Graph y = 2∛(x + 1) − 3: its center, and y at x = 7” or “Find x where y = 5.”
Rational exponents
Use this for “Evaluate 82/3”, “Graph y = 2x−1/2” or “Solve x3/2 = 27.”