Surds equations
Yue Kwok Choy
(1)
(a)
Squaring creates roots.
(b) Beware of hidden givens.
(c)
Check your roots.
Example 1
3x 2  4 x  5  2 x  3
Solve :
Solution
for real roots.
Squaring the given equation, we have:
3x 2  4x  5  2x  32

2
 3x  4x  5  0

2x  3  0

Note that (2) and (3)
x = 4 3

are hidden givens .
x2 – 8x + 4 = 0 
From (1),
(1)
(2)
(3)
does not satisfy (3)
x  4 3
since


2x  3  2 4  3  3  5  4 3  0
x = 4  3 is rejected.
x = 4 3
satisfy both (2) and (3) .
(Ans) x = 4  3
is the only root.
Example 2
Solve :
x
Solution
2

2
 x  x 2 1  0
for real roots.
From the given equation :
x 2  x  0 (1)
 2
 x  1  0 (2)
Solving, the only root is x = – 1 .
(2)
In some surds equations, transpose term before squaring may be better.
Example 3
Solve :
Solution
2x  1  2x  1  2
for real roots.
Transpose term in the given equation :
2x  1  2  2x  1
2x  1  4  4 2x  1  2x  1
Squaring,
4 2x  1  2
2 2x  1  1
Squaring again, 4(2x – 1) = 1

Exercise
x=
Solve
5
8
, which is a good root on checking .
4x  3  4x  1  2
for real roots.
Ans :
1
.
4
1
Example 4
Solve :
Solution
5x  1  x  2
for real roots.
Transpose term in the given equation :
5x  1  2  x .
Squaring, 5x  1  x  4 x  4
4 x  4x  3
16x = 16x2 – 24x + 9
Squaring again,
16x2 – 40x + 9 = 0
(4x – 9) (4x – 1) = 0

(3)
x  1  1  2x  1
Solve
9
4
1
4
or
1
is a redundant root and should be rejected.
4
On checking, x 
Exercise
x
for real roots.
Ans :

x
9
4
2 7 1
9
Conjugate method
Example 4
Solve :
3x  4  3x  6  10
Solution
Let
for real roots.
3x  4  3x  6  10
…. (1)
3x  4  3x  6  y
…. (2)
3x  4  3x  6  10 y
(1)(2),

y=1.
3x  4  3x  6  1
From (2),
(1) + (3),
2 3x  4  11
Squaring,
4(3x + 4) = 121

x=
35
.
4
…. (3)
(on checking, this is a good root)
Example 5
Solve :
x 2  3x  7  x 2  3x  9  2
Solution
Let
for real roots.
x 2  3x  7  x 2  3x  9  2
…. (1)
x 2  3x  7  x 2  3x  9  y
…. (2)
x
(1)(2),

From (2),
[(1) + (3)]/2,
2
 

 3x  7  x 2  3x  9  2y
y=8.
x 2  3x  7  x 2  3x  9  8
x 2  3x  7 = 5
Squaring and simplify,
…. (3)
…. (4)
x2 + 3x – 18 = 0
2
(x – 3) (x + 6) = 0

x=3,–6
Since both sides of (4) are positive, we do not have any redundant root on squaring.
Exercise
(4)
x 2  4x  34  x 2  4x  11  9
Solve
for real roots.
Ans :
3,
5
3
Change of variables
Example 6
Solve :
x 2  2x  2x 2  4x  3 for real roots.
Solution
x 2  2x  2x 2  4x  3
Put
then
…. (1)
…. (2)
y = x2 – 2x
2x  4x  3 = 2y + 3
2
The given equation then becomes
Squaring ,
y2 – 2y – 3 = 0,
y  2y  3
…. (3)
(y – 3)(y + 1) = 0
Since from (3),
y  0,
y + 1  0,
 y=3
2
(4)(2),
x – 2x – 3 = 0 ,
(x – 3)(x + 1) = 0
 x = 3, –1 .
…. (4)
Example 7
Solve :
Solution
x  2  x  k  2  1 , where
Let
k is a constant .
 u  x  2  0

v  x  k  2  0
Then the given equation becomes
 u  v 1
:  2
2
u  v  k

u  v 1
u 

u  v  1



u  v u  v   k
u  v  k
v 

Since
therefore
Exercise
Solve
x 1 
1
k  1  0,
2
1
k  1
2
1
k  1
2
x  k 1 
1
k  1  0
2
1

x  k  12  2 , when k  1

4

, when k  1
 no solution
x 1 
x 1
1
x
Ans: x 
1 5
2
3
4