Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.53

Example 89 : 2x^2 – 7x + 6 = 0
L.H.S. 2x^2 – 3x – 4x + 6 = x (2x – 3) – 2 (2x – 3) = (2x – 3) (x – 2)
Now, (2x – 3) (x – 2) = 0 is true if either 2x – 3 or, x – 2 = 0
From 2x – 3 = 0 we get 2x = 3 or x = 3/2, and from x –2 = 0 we get x = 2
Hence the values (or roots) are 3/2, 2.
By the. Second method (b) for ax^2 + bx + c = 0, we have

b b 4ac^2

x=− ± 2a− , (the proof is not shown at present)

Example 90 : Solve 2x^2 - 7x + 6 = 0 Here a = 2, b = -7, c = 6

(7) ( 7) 4.2.6^2 7 49 48 7 1 8 6 3

∴ =x − − ± 2.2− − = ± 4 − = 4 4 4 2± = , 2,=

Example 91 : Solve x 27 x 11 (^2) −^9 = (^225) −
Cross multiplying, 25x^2 - 675 = 9x^2 - 99, or, 16x^2 = 576 or, x^2 = 36 or, x = ± 6
Example 92 : Solve x^2 -7x+12 = 0
Solution :
This can be expressed as x^2 -3x - 4x + 12 = 0 or, x (x - 3) - 4 (x - 3) = 0
or, (x - 3) (x - 4) = 0 Hence, x = 3, 4.
Alternatively, x=− ±7 49 48 7 1 2 − = 2 ±. Hence, x = 4, 3. (Here, a = 1, b = - 7, c = 12)
Equation reducible to Quadratics :
There are various types of equations, not quadratic in form, which can be reduced to quadratic forms by
suitable transformation, as shown below :
Example 93 : Solve x^4 - 10x^2 + 9 = 0.
Solution :
Taking, x^2 = u, we get u^2 - l0u + 9 = 0
or, (u - 9) (u - 1) = 0 ; either (u - 9) = 0 or, (u - 1) = 0 Hence, u = 9, 1.
When u = 9, x^2 = 9 or, x = ± 3
Again, u = 1, x^2 = 1 or, x = ± 1 Hence, x = ± 3, ±1.
Here the power of x is 4, so we get four values of x.
Example 94 : Solve : (1 + x)1/3 + (1 - x)1/3 = 21/3
Solution :
We get, (1 + x) + (1 - x) + 3 (1 + x)1/3 (1 - x)1/3. [(1 + x)1/3 + (1 - x)1/3] = 2 (cubing sides).
or, 2 + 3(1 - x^2 )1/3.21/3 = 2 or, 3(1 - x^2 )1/3 .21/3 = 0
or, (1-x^2 )1/3 = 0, as 3.21/3 ≠ 0
or, 1 - x^2 = 0 (cubing again)
or, x^2 =1 ∴ x = ± 1