Maths, Maths and only Maths

1. EQUATION REDUCIBLE TO
QUADRATIC FORM
PRESENTED BY: Mahrukh Shehzadi

3. REDUCIBLE EQUATIONS
.
There are some types of
equations, which under some
proper substitution cab be
reduced into quadratic form or
equation.
Example:
In this method we have to reduce some equations into quadratic form and
then we will find the value of x.
Let us understand this with the help of a simple example:

4. Let us suppose this example:
𝟐𝒙 𝟒 − 𝟗𝒙 𝟐 + 𝟒 = 𝟎
 Putting 𝒙 𝟐
= 𝒚
𝑥4
= 𝑥2 2
= 𝑦2
𝐢𝐧 𝐞𝐪. 𝐢 .
Thus,
2𝑦2
− 9𝑦 + 4 = 0
𝒂 = 2 , 𝒃 = −9 , 𝒄 = −4
 By putting quadratic formula:
𝒚 =
−𝒃 ± 𝒃 𝟐 − 𝟒𝒂𝒄
𝟐𝒂
𝒚 =
− −𝟗 ± −𝟗 𝟐 − 𝟒 𝟐 𝟒
𝟐 𝟐

5. 𝒚 =
𝟗 ± 𝟖𝟏 − 𝟑𝟐
𝟒
𝒚 =
𝟗 ± 𝟒𝟗
𝟒
𝒚 =
𝟗+𝟕
𝟒
⇒ 𝒚 =
𝟏𝟔
𝟒
𝒚 =
𝟗−𝟕
𝟒
⇒ 𝒚 =
𝟐
𝟒
𝒚 = 𝟒 𝒚 =
𝟏
𝟐
Putting 𝒚 = 𝒙 𝟐:
𝒙 𝟐 = 𝟒 𝒙 𝟐 =
𝟏
𝟐
𝒙 𝟐 = 𝟐 𝟐 𝒙 𝟐 =
𝟏
𝟐
𝒙 = ±𝟐 𝒙 = ±
𝟏
𝟐
Solution set: 𝑥 = ±2, ±
1
2

6. Types
Type (i)
Type(ii)
Type (iii)Type (iv)
Type (v)

7. TYPE (1)
The equation of the types:
𝒂𝒙 𝟒 − 𝒃𝒙 𝟐 + 𝒄 = 𝟎
Example:
Replacing 𝒙 𝟐 = 𝒚 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝒂𝒙 𝟒 + 𝒃𝒙 𝟐 + 𝒄 = 𝟎,
we get a quadratic equation in y.
Let us suppose this example:
2𝒙 𝟒
− 𝟏𝟏𝒙 𝟐
+ 𝟓 = 𝟎

8.  Putting 𝒙 𝟐
= 𝒚
𝑥4
= 𝑥2 2
= 𝑦2
𝐢𝐧 𝐞𝐪. 𝐢 .
Thus,
𝟐𝒚 𝟐
− 𝟏𝟏𝒚 + 𝟓 = 𝟎
𝒂 = 2 , 𝒃 = −11 , 𝒄 = 5
 By putting quadratic formula:
𝒚 =
−𝒃 ± 𝒃 𝟐 − 𝟒𝒂𝒄
𝟐𝒂
𝒚 =
− −𝟏𝟏 ± −𝟏𝟏 𝟐 − 𝟒 𝟐 𝟓
𝟐 𝟐
𝒚 =
𝟏𝟏 ± 𝟏𝟐𝟏 − 𝟒𝟎
𝟒
𝒚 =
𝟏𝟏 ± 𝟖𝟏
𝟒
2𝒙 𝟒 − 𝟏𝟏𝒙 𝟐 + 𝟓 = 𝟎

9. 𝒚 =
𝟏𝟏+𝟗
𝟒
⇒ 𝒚 =
𝟐𝟎
𝟒
𝒚 =
𝟏𝟏−𝟗
𝟒
⇒ 𝒚 =
𝟐
𝟒
𝒚 = 𝟓 𝒚 =
𝟏
𝟐
Putting 𝒚 = 𝒙 𝟐:
𝒙 𝟐
= 𝟓 𝒙 𝟐
=
𝟏
𝟐
𝒙 𝟐 = 𝟓 𝒙 𝟐 =
𝟏
𝟐
𝒙 = ± 𝟓 𝒙 = ±
𝟏
𝟐
Solution set: 𝑥 = ± 𝟓, ±
1
2

10. TYPE (2)
The equation of the type:
𝒂𝒑 𝒙 +
𝒃
𝒑 𝒙
= 𝒄
Example:
In this types of equation, the 𝑥 𝑖𝑛 𝑎𝑝 𝑥 +
𝑏
𝑃 𝑥
𝑖𝑠 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑦.
 Let us suppose this example:
 2𝑥2 + 1 +
3
2𝑥2+1
= 4

11.  Putting 2𝑥2
+ 1 = 𝑥 = 𝑦:
𝒚 +
𝟑
𝒚
= 𝟒
𝒚 𝟐 + 𝟑 − 𝟒𝒚 = 𝟎
𝒚 𝟐
− 𝟒𝒚 + 𝟑 = 𝟎
 By putting quadratic formula: y=
−𝒃± 𝒃 𝟐−𝟒𝒂𝒄
𝟐𝒂
𝒚 =
− −𝟒 ± −𝟒 𝟐 − 𝟒 𝟏 𝟑
𝟐 𝟏
𝒚 =
𝟒 ± 𝟏𝟔 − 𝟏𝟐
𝟐
𝒚 =
𝟒 ± 𝟒
𝟐
 2𝑥2 + 1 +
3
2𝑥2+1
= 4

12. 𝒚 =
𝟒+𝟐
𝟐
⇒ 𝒚 =
𝟔
𝟐
𝒚 =
𝟒−𝟐
𝟐
⇒ 𝒚 =
𝟐
𝟐
𝒚 = 𝟑 𝒚 = 𝟏
Putting 2𝑥2
+ 1 = 𝑦:
2𝑥2 + 1 = 𝟑 2𝑥2 + 1 = 𝟏
2𝑥2
= 3 − 1 2𝑥2
= 1 − 1
2𝑥2 = 2 2𝑥2 = 0
𝑥2 = 1 𝑥2 = 0
𝑥 = 0,1Solution set:

13. TYPE (3)
The type of equation
𝒂 𝒙 𝟐
+
𝟏
𝒙 𝟐
+ 𝒃 𝒙 +
𝟏
𝒙
+ 𝑪 = 𝟎
Example:
The equations are called as reciprocal equations. An equation is said to be
reciprocal if it remains unchanged when 𝒙 𝑖𝑠 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦
𝟏
𝒙
.
This type can also be written as :
𝒂𝒙 𝟒 + 𝒃𝒙 𝟑 + 𝒄𝒙 𝟐 + 𝒃𝒙 + 𝒂 = 𝟎
In this type we have to convert this equation into the square of x.

14.  Let us suppose this example:
𝑥4 − 2𝑥3 − 2𝑥2 + 2𝑥 + 1 = 0
Diving the equation with 𝑥2:
𝑥4
𝑥2
+
2𝑥3
𝑥2
−
2𝑥2
𝑥2
+
2𝑥
𝑥2
+
1
𝑥2
= 0
𝑥2
− 2𝑥 − 2 +
2
𝑥
+
1
𝑥2
= 0
𝑥2
+
1
𝑥2
− 2 𝑥 −
1
𝑥
− 2 = 0
Let: 𝑥 −
1
𝑥
= 𝑦 :
𝑥 −
1
𝑥
2
= 𝑦2
⇒ 𝑥2
+
1
𝑥2
− 2 = 𝑦2

15. 𝑥2
+
1
𝑥2 = y2
+ 2
By putting the values:
𝑦2 + 2 − 2 𝑦 − 2 = 0
𝑦2
+ 2 − 2𝑦 − 2 = 0
𝑦2 − 2𝑦 = 0
𝑦 𝑦 − 2 = 0
𝑦 = 0 𝑦 = −2
By putting the values 𝑥 −
1
𝑥
= 𝑦:
𝑥 −
1
𝑥
= 0 𝑥 −
1
x
= 2
𝑥2
− 1 = 𝑂 𝑥2
− 1 = 2𝑥
𝑥2 = 1 𝑥2 − 2𝑥 − 1 = 0
𝑥 = ±1 𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎

16. 𝑥 =
−𝒃± 𝒃 𝟐−𝟒𝒂𝒄
𝟐𝒂
𝑥 =
− −𝟐 ± −𝟐 𝟐−𝟒 𝟏 −𝟏
𝟐 𝟏
𝑥 =
𝟐 ± 𝟒 + 𝟒
𝟐
𝑥 =
2± 2×𝟒
2
𝑥 =
2±2 2
2
𝑥 = 𝟐
𝟏±𝟏 2
2
𝑥 = 1 ± 𝟏 2
𝑥 = +1, ±𝟏 2
𝑥 = ±1
Solution set:

17. TYPE (4)
The type of equation:
Exponential equation
Example:
In this type of equation or in exponential equation variable occur in exponents.
e.g. the variable 𝑥 occur in exponent like 8 𝑥 .
Let us under stand this type with the help of an example:

18. 51+𝑥 + 51−𝑥 = 26
5.5 𝑥 + 5.5−𝑥 = 26
5.5 𝑥 +
5
5 𝑥
− 26 = 0
Let 5 𝑥 = 𝑦:
5𝑦 +
5
𝑦
− 26 = 0
5𝑦2 + 5 − 26𝑦 = 0
5𝑦2 − 26𝑦 + 5 = 0
By factorization:
5𝑦2 − 25𝑦 − 𝑦 + 5 = 0
5𝑦 𝑦 − 5 − 1 𝑦 − 5 = 0
𝑦 − 5 5y − 1 = 0
𝑦 − 5 = 0 5𝑦, 1 = 0

19. 𝑦 − 5 = 0 5𝑦 − 1 = 0
𝑦 = 5 𝑦 = 1
5
Putting 𝑦 = 5 𝑥:
5 𝑥
= 5 5 𝑥
= 1
5
𝑥 = 1 5 𝑥 = 5−1
𝑥 = −1
𝑥 = 1, −1 𝑜𝑟 ±1Solution set:

20. TYPE (5)
The type of the equation:
𝒙 + 𝒂 𝒙 + 𝒃 𝒙 + 𝒄 𝒙 + 𝒅 = 𝒌
Example:
In this type of equation the 𝒂 + 𝒃 = 𝒄 + 𝒅
First we will multiply two brackets and then place y in the place of same
terms.

21.  𝑥 − 1 𝑥 − 2 𝑥 − 8 𝑥 + 5 = −360
𝑥2
− 2𝑥 − 𝑥 + 2 𝑥2
− 8𝑥 + 5𝑥 − 40 = −360
𝑥2
− 3𝑥 + 2 𝑥2
− 3𝑥 − 40 = −360
Let 𝑦 = 𝑥 − 3𝑥:
𝑦 + 2 𝑦 − 40 = −360
𝑦2
− 40y +2𝑦 − 80 + 360 = 0
𝑦2 − 38y + 280 = 0
𝑦2 − 28y − 10𝑦 + 280 = 0
𝑦 − 28 𝑦 − 10 = 0
𝑦 − 28 = 0 𝑦 − 10 = 0
𝑦 = 28 𝑦 = 10
Putting 𝑦 = 𝑥 − 3𝑥:
𝑥2 − 3𝑥 = 28 𝑥2 − 3𝑥 = 10
𝑥2 − 3𝑥 − 28 = 0 𝑥2 − 3𝑥 − 10 = 0

22. 𝑥2 − 7𝑥 + 4𝑥 − 28 = 0 𝑥2 − 5𝑥 + 2𝑥 − 10 = 0
𝑥 𝑥 − 7 + 4 𝑥 − 7 = 0 𝑥 𝑥 − 5 + 2 𝑥 − 5 = 0
𝑥 + 4 𝑥 − 7 = 0 𝑥 + 2 𝑥 − 5 = 0
𝑥 + 4 = 0 𝑥 − 7 = 0 𝑥 + 2 = 0 𝑥 − 5 = 0
x= −4 𝑥 = 7 𝑥 = −2 𝑥 = 5
Solution set: 𝑥 = −4, −2,5,7

23. ANY
QUESTION?
THANK YOU 