Q9 - Solve the Following equations (x - 1)(x - 2)(x - 8)(x + 5) + 360 = 0 - Fsc Part 1 Mathematics (Complete Solution) [ Notes, Key book, Solved Exercises, Solved Questions ]

Solutions

As

-1 - 2 - 8 + 5

2

=

-6

2

= -3, So we can write as

[(x - 1)(x - 2)] [(x - 8)(x + 5)] + 360 = 0

(x² - 2x - x + 2)(x² + 5x - 8x - 40) + 360 = 0

(x² - 3x + 2)(x² - 3x - 40) + 360 = 0

(x² - 3x + 2)(x² - 3x + 2 - 42) + 360 = 0

Let x² - 3x + 2 = t and Put in the equation

(t)(t - 42) + 360 = 0

t² - 42t + 360 = 0

Using the quadratic formula, we have

t =

-(-42) ±

√	(-42)² - 4(1)(360)

2(1)

t =

42 ±

√	1764 - 1440

2

t =

42 ±

√

324

2

t =

42 ±

18

2

Either

t =

42 -

18

2

OR

t =

42 +

18

2

t =

24

2

t =

60

2

t = 12

t = 30

By Letting t = x² - 3x + 2, Putting the value of t, we get

x² - 3x + 2 = 12

x² - 3x + 2 = 30

x² - 3x + 2 - 12 = 0

x² - 3x + 2 - 30 = 0

x² - 3x - 10 = 0 -------------------- (i)

x² - 3x - 28 = 0 -------------------- (ii)

By Equation (i) x² - 3x - 10 = 0

Using the Quadratic formula, we have

x =

-(-3) ±

√	(-3)² - 4(1)(-10)

2(1)

x =

3 ±

√

9 + 40

2

x =

3 ±

√

49

2

x =

3 ±

7

2

Either

x =

3 -

7

2

OR

x =

3 +

7

2

x =

-4

2

x =

10

2

x =

-2

x =

5

By Equation (ii) x² - 3x - 28 = 0

Using the Quadratic formula, we have

x =

-(-3) ±

√	(-3)² - 4(1)(-28)

2(1)

x =

3 ±

√

9 + 112

2

x =

3 ±

√

121

2

x =

3 ±

11

2

Either

x =

3 -

11

2

OR

x =

3 +

11

2

x =

-8

2

x =

14

2

x =

-4

x =

7

Hence the Solution Set of given equation is {-4, -2, 5, 7}

Q9
Solve the Following equations (x - 1)(x - 2)(x - 8)(x + 5) + 360 = 0

Other Topics

Q9 Solve the Following equations (x - 1)(x - 2)(x - 8)(x + 5) + 360 = 0

Other Topics

Q9
Solve the Following equations (x - 1)(x - 2)(x - 8)(x + 5) + 360 = 0