Let us take the general form of a quadratic equation:

ax² + bx + c = 0            (1)



Since a ≠ 0, hence multiplying both sides of (1) by 4a we get,

4a²x² + 4abx + 4ac = 0 or (2ax)² + 2.2ax.b + b² – b² + 4ac = 0

Or, (2ax + b)² = b² – 4ac

2ax + b = b² – 4ac

x =

Hence, the roots of (1) are

Let, = and ß =

Hence, + ß =    +   

Or + ß =   =      -   b/a   =   -                  (2)

Again ß = x

Or ß = =

Or ß = = =               (3)

Example 1:

If the roots of the equation 2x² - 9x - 3 = 0 be and ß, then find + ß and ß.

Solution:

We know that + ß = - = - =

And ß = =               (Answer)

Example 2:

If one root of the quadratic equation x² – x - 1 = 0 is a, prove that its other root is ³ - 3.

Solution:

x² – x - 1 = 0               (1)

Let ß be the other root of the equation (1). Then,

+ ß = = 1 or ß = 1 -

Since is a root of the equation (1) hence, ² – - 1 = 0 or ² = + 1

Now, ³ - 3 = . ² - 3 = ( + 1) - 3       [Since ² = + 1]

                           = ² + - 3 = + 1 - 2 = 1 – = ß       [Since ß = 1 – ]

Hence, the other root of equation (1) is ³ - 3.                (Proved)

Example 3:

If a² = 5a – 3 and b² = 5b – 3, (a ≠ b), find the quadratic equation whose roots are and .

Solution:



Hence, a + b = - = 5 and ab = = 3.

Now, the sum of the roots of the required equation

= + = = = = =

And the product of the roots of the required equation = . = 1.

Hence, the required equation is x² – x + 1 = 0 or 3x² – 19x + 3 = 0.                (Answer)

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