Grade 11 Math · Problem type

Solve a rational equation

Use this for “Solve x/(x − 2) = 2/(x − 2) + 3/x” and reject the extraneous solution.

Chart

f(x) = (x − 3)(x − 2)x(x − 2)−2−101234−8−6−4−20246810xyx = 0y = 1(3, 0)
Zero: x = 3
Vertical asymptote x = 0
Horizontal asymptote y = 1
Hole at (2, −0.5)
This is an example. Type your own values to start a new problem.
x
x
−
=
a
x
−
+
b
x
x₁
Larger candidate
worked out (not typed)
x₂
Smaller candidate
worked out (not typed)

Formulas

x₁
=
((
a
+
b
)
+
√(
(
a
+
b
)
2
−
4
×
b
×
p
))
÷
2
3
=
((2
+
3)
+
√(
(2 + 3)
2
−
4
×
3
×
2))
÷
2
x₂
=
((
a
+
b
)
−
√(
(
a
+
b
)
2
−
4
×
b
×
p
))
÷
2
2
=
((2
+
3)
−
√(
(2 + 3)
2
−
4
×
3
×
2))
÷
2

Assumptions

•
Multiply every term by x(x − p) to clear the fractions: x² − (a + b)x + bp = 0.
•
A candidate that makes a denominator 0 (x = 0 or x = p) is extraneous: reject it.
•
On the graph of the left side minus the right, an extraneous candidate is a hole.

Step-by-step

WE KNOW
p = 2, a = 2, b = 3
FIND
larger candidate (x₁), smaller candidate (x₂)
Step 1 · Find larger candidate (x₁)
Use x₁ = ((a + b) + √((a + b)² − 4 × b × p)) ÷ 2
Multiply
by
x(x
−
p
),
then
use
the
quadratic
formula.
x₁
=
((2
+
3)
+
√(
(2 + 3)
2
−
4
×
3
×
2))
÷
2
x
2
−
5x
+
6
=
0
x₁
=
5 +
√
1
2
x₁
=
5 + 1
2
x
=
3:
3
1
=
2
1
+
3
3
✓
x₁
=
3
Step 2 · Find smaller candidate (x₂)
Use x₂ = ((a + b) − √((a + b)² − 4 × b × p)) ÷ 2
The same formula with the minus sign.
x₂
=
((2
+
3)
−
√(
(2 + 3)
2
−
4
×
3
×
2))
÷
2
x₂
=
5 −
√
1
2
x₂
=
5 − 1
2
x = 2 makes x − 2 = 0: reject it
x₂
=
2
Check
Put the numbers back in. Both sides should match.
3
=
((2
+
3)
+
√(
(2 + 3)
2
−
4
×
3
×
2))
÷
2
✓
2
=
((2
+
3)
−
√(
(2 + 3)
2
−
4
×
3
×
2))
÷
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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