Page 12 - 10-Math-1 QUADRATIC EQUATIONS

11. .QQuuaaddrraatitcicEEqquuaatitoionnss eeLleeaarrnn..Ppuunnjjaabb

Example 4: Solve the equation 51+x + x + 51 - x = 26.

Solution: 51+x + x + 51 - x = 26 Let 5x = y.
51 . 5x + 51 . 5-x = 26 or

Then equation (i) becomes 5y + 5 - 26 = 0
y

5y2 + 5 - 26y = 0 or 5y2 - 26y + 5 = 0

5y2 - 25y - y + 5 = 0

5y (y - 5) - 1(y - 5) = 0

(y - 5) (5y - 1) = 0
Either y - 5 = 0 or 5y - 1 = 0 , that is, y = 5 or 5y = 1 ⇒ y = 1
5
Put y = 5x

5x = 51 or 5x = 5-1 x = 1 or x = -1

∴ The solution set is { +1}.

Type (v) The equations of the type:
(x + a) (x + b) (x + c) (x + d) = k , where a + b = c + d

Example 5: Solve the equation (x - 1) (x + 2) (x + 8) (x + 5) = 19.

Solution: (x - 1) (x + 2) (x + 8) (x + 5) = 19 ( a -1 + 8 = 2 + 5)

or [(x - 1) (x + 8)] [(x + 2) (x + 5)] - 19 = 0
(x2 + 7x - 8) (x2 + 7x + 10) - 19 = 0 (i)
Let x2 + 7x = y

Version 1.1 Then eq. (i) becomes (y - 8) (y + 10) - 19 = 0
y2 + 2y - 80 - 19 = 0
y2 + 2y - 99 = 0
y2 + 11y - 9y - 99 = 0
y(y + 11) - 9(y + 11) = 0
(y + 11) (y - 9) = 0
Either y + 11 = 0 or y - 9 = 0
Put y = x2 + 7x, so
x2 + 7x + 11 = 0 or x2 + 7x - 9 = 0

12