Important Key Points

Example 3.23
Solve the equation
(x − 2) (x − 7) (x − 3) (x + 2) +19 = 0 .

Solution :
Rewriting the equation as
(x − 2) (x − 3) (x − 7) (x + 2) +19 = 0.
the given equation becomes
($x^2$ − 5x + 6) ($x^2$ − 5x −14) +19 = 0.
If we take y=$x^2-5x$,
Then the equation becomes
( y + 6) ( y −14) +19 = 0.
$y^2+6y-14y-84+19=0$
$y^2-8y-65=0$
This is a quadratic equation in terms of 'y',
we can write it as (y-13)(y+8)=0 by factorization method.
y=13, y=-8.
Since we have taken y=$x^2-5x$, we get two quadratic equation as $x^2$ − 5x −13 = 0 and $x^2$ − 5x + 5 = 0, These two equations are quadratic equations in terms of 'x'. By solving this in formula method [x=$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$], we get $x=\frac{5\pm\sqrt{77}}{2}, x=\frac{5\pm\sqrt{5}}{2}$

Example 3.24
Solve the equation
(2x-3)(6x-1)(3x-2)(x-2)-5 = 0.

Solution :
Rewriting the equation as
(2x-3) (3x-2) (6x-1) (x-2)-5 = 0
the given equation becomes
($6x^2$ −13x+6) ($6x^2$ −13x+2)−5 = 0 .
If we take y=$6x^2 −13x$,
Then the equation becomes
( y+6) ( y+2)−5 = 0
$y^2+6y+2y+12=0$
$y^2$ +8y+7 = 0 .
This is a quadratic equation in terms of 'y',
we can write it as (y+1)(y+7)=0 by factorization method.
y=-1, y=-7.
Since we have taken y=$6x^2 −13x$, we get two quadratic equation as
$6x^2$ −13x+1 = 0 and $6x^2$ −13x+7 = 0. These two equations are quadratic equations in terms of 'x'. By solving this we get x=1, x=$\frac{7}{6}$, x=$\frac{13\pm\sqrt{145}}{12}$

Exercise 3.4

1. Solve
(i) (x – 5) (x – 7) (x + 6) (x + 4) = 504
(ii) (x – 4) (x – 7) (x – 2) (x + 1) = 16

Solution :
(i) Rewriting the equation as
(x – 5) (x + 4) (x – 7) (x + 6) – 504 = 0
the given equation becomes,
(x² – x – 20) (x² – x – 42) – 504 = 0
Put y = x² – x
Then the equation becomes
(y – 20) (y – 42) – 504 = 0
y² – 42y – 20y + 840 – 504 = 0
y² – 62y + 336 = 0
This is a quadratic equation in terms of 'y',
(y – 56) (y – 6) = 0 by factorization method.
y = 56 or y = 6
Since we have taken y=$x² – x$, we get two quadratic equation as
x² – x = 56 or x² – x = 6
x² – x – 56 = 0 or x² – x – 6 = 0
These two equations are quadratic equations in terms of 'x'. By solving this by factorization method we get,
(x – 8)(x + 7) = 0 or (x + 2)(x – 3) = 0
x = 8, x=-7, x = 3, x = -2

(ii)Rewriting the equation as (x – 4)(x – 2)(x- 7)(x + 1) – 16 = 0
the given equation becomes,
(x² – 6x + 8) (x² – 6x – 7) – 16 = 0
Put y = x² – 6x
Then the equation becomes
(y + 8) (y + 7) – 16 = 0
y² – 7y + 8y – 56 – 16 = 0
y² – y – 72 = 0
This is a quadratic equation in terms of 'y',
(y + 9) (y – 8) = 0 by factorization method.
y = -9 or y = 8
Since we have taken y=$x² – 6x$, we get two quadratic equation as

If y = -9 ⇒ x² – 6x = -9
x² – 6x + 9 = 0
(x – 3)² = 0
x = 3, x=3
If y = 8 ⇒ x² – 6x = 8
x² – 6x – 8 = 0
Solving this by formula method [x=$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$],
Here a=1, b= -6, c=-8
x=$\frac{6\pm\sqrt{(-6)^2-4(1)(-8)}}{2(1)}$
x=$\frac{6\pm\sqrt{36+32}}{2}$
x=$\frac{6\pm\sqrt{68}}{2}$
x=$\frac{6\pm2\sqrt{17}}{2}$
x=2($\frac{3\pm2\sqrt{17}}{2}$)
x=$3\pm\sqrt{17}$
Finally the four values of x are x = 3, x=3, x=$3+\sqrt{17}$, x=$3-\sqrt{17}$

2. Solve :
(2x – 1) (x + 3) (x – 2) (2x + 3) + 20 = 0.

Solution :
Rewriting the equation as
(2x – 1) (2x + 3) (x + 3) (x – 2) + 20 = 0
the given equation becomes
⇒ (4$x^2$ + 6x – 2x – 3) ($x^2$ – 2x + 3x – 6) + 20 = 0
⇒ (4$x^2$ + 4x – 3) ($x^2$ + x – 6) + 20 = 0
⇒ [4($x^2$ + x) – 3] [$x^2$ + x – 6] + 20 = 0
Let y = $x^2$ + x
Then the equation becomes
⇒ (4y – 3) (y – 6) + 20 = 0
⇒ 4$y^2$ – 24y – 3y + 18 + 20 = 0
⇒ 4$y^2$ – 27y + 38 = 0
This is a quadratic equation in terms of 'y',
we can write it as (4y – 19) (y – 2) = 0 by factorization method.
(4y – 19) = 0
4($x^2$+ x) – 19 = 0
4$x^2$ + 4x – 19 = 0
Here a=4, b=4, c=-19.
x=$\frac{6\pm\sqrt{(4)^2-4(4)(-19)}}{2(4)}$
x=$\frac{-4\pm\sqrt{16+304}}{8}$
x=$\frac{-4\pm\sqrt{320}}{8}$
x=$\frac{-4\pm8\sqrt{5}}{8}$
x=4($\frac{-1\pm2\sqrt{5}}{8}$)
x=$\frac{3\pm\sqrt{17}}{2}$
or
(y – 2) = 0
x2 + x – 2 = 0
(x + 2) (x – 1) = 0 by factorization method.
x = -2, x=+1
The four roots of the given equations are x=$\frac{3+\sqrt{17}}{2}$, x=$\frac{3-\sqrt{17}}{2}$, x = -2, x=+1