Page 14 - 11-Math-4 Quadratic Equations

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x - 3 =0 ⇒ x-3 = 0 ⇒x = 3 6x +13 =0

Now solve the equation x + 7 + x + 2 -

⇒ x + 7 + x + 2= 6x +13

⇒ x + 7 + x + 2 + 2 (x + 7)(x + 2) = 6x +13 (Squaring both sides)

⇒ 2 (x + 7)(x + 2) = 4x + 4

⇒ x2 + 9x + 14 = 2x + 2
⇒ x 2 + 9x + 14 = 4x 2 + 8x + 4 (Squaring both sides again)
⇒ 3x2 - x - 10 = 0
⇒ (3x + 5)(x - 2 ) = 0

⇒ x = - 5 ,2
3

Thus possible roots are 3, 2, - 5 .
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On verification, it is found that - 5 is an extraneous root. Hence solution set = {2, 3}

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iv) The Equations of the form: ax2 + bx + c + px2 + qx + r = mx + n
where, (mx + n) is a factor of (ax2 + bx + c) - (px2 + qx + r)

Example 4: Solve the equation: 3x2 - 7x - 30 - 2x2 - 7x - 5 =x - 5

Solution: Let 3x2 =- 7x - 30 a and 2x=2 - 7x - 5 b
Now a2 - b2 = ( 3x2 - 7x - 30 ) - (2x2 - 7x - 5)
a2 - b2 = x2 -25 (i)
The given equation can be written as:
a - b = x - 5 (ii)

(a + b)(a - b) = (x + 5)(x - 5) [From (i) and (ii)]
a-b x - 5

⇒ a + b = x + 5 (iii)

2a = 2x [From (ii) and (iii)]

⇒ a=x

version: 1.1

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