How do you solve the Relationship between the coefficient and roots of quadratic equations?

Answer 1

TO FIND THE RELATION BETWEEN ROOTS AND COEFFICIENTS OF A QUADRATIC EQUATION:

Let us take the general form of a quadratic equation:

ax2 + bx + c = 0 (1)

where a(≠ 0) is the coefficient of x2 , b is the coefficient of x and c is a constant term. If and ß be the roots of the equation, then we have to find the relations of and ß with a, b and c.

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)

Equations (2) and (3) represent the required relations between roots (that is, and ß) and coefficients (that is, a, b and c) of equation (1).

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:

Given (a ≠ b) and a2 = 5a - 3 and b2 = 5b - 3, hence it is clear that a and b are the roots of the equation x2 = 5x - 3 or x2 - 5x + 3 = 0.

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)