Page 11 - 10-Math-1 QUADRATIC EQUATIONS
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which is simplified as a - bx + cx2 - bx3 + ax4 = 0. We get the same
equation. Thus ax4 - bx3 + cx2 - bx + a = 0 is a reciprocal equation.
The method for solving reciprocal equation is illustrated through
an example.
Example 3: Solve the equation 2x4 - 5x3 - 14x2 - 5x + 2 = 0.
Solution: 2x4 - 5x3 - 14x2 - 5x + 2 = 0
Dividing each term by x2
(i)
So equation (i) becomes
2(y2 - 2) - 5y - 14 = 0 or 2y2 - 4 - 5y - 14 = 0
2y2 - 5y - 18 = 0
2y2 - 9y + 4y - 18 = 0 or y(2y - 9) + 2(2y - 9) = 0
⇒ (2y - 9) (y + 2) = 0
Either 2y - 9 = 0 or y + 2 = 0
As y= x + 1 , so we have
x
= 2ï£«ï£¬ï£ x + 1x  - 9 0
or =x + 1 + 2 0
x
=2x2 - 9x + 2 0 or=x2 + 2x + 1 0
2x2 - 9x + 2 = 0 or x2 + 2x + 1 = 0
Type (iv) Exponential equations: Version 1.1
In exponential equations, variable occurs in exponent. The
method of solving such equations is illustrated through an example.
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