Equation From Roots Solution3

In this page equation from roots solution3 we are going to see solution for the worksheet framing quadratic equation from roots.

Find the zeroes of the following quadratic polynomials and verify the basic relationship between the zeroes and coefficients

(v) x²  - 15

First we have to compare the given equation with the general form of

a x² + b x + c

Let p(x) = x²  - 15

So, p(x) = 0

x²  - 15 = 0

x² = 15

x = √15

x = √15     x = -√15

p (√15) = x2  - 15

= (√15) 2 - 15

= 15 - 1 5

= 0

p (-√15) = x²  - 15

= (-√15)² - 15

= 15 - 1 5

= 0

Hence the zeroes of p(x) are √15 and -√15

Thus, Sum of zeroes = 0 and the product of zeroes = -15

From the basic relationships, we get

The sum of the zeroes = -coefficient of x/coefficient of x²

= 0/1

= 0

The product of the zeroes = constant term/coefficient of x²

= -15/1

= -15

Thus the basic relationship verified.

(vi) 3 x² - 5 x + 2

First we have to compare the given equation with the general form of a x² + b x + c

Let p(x) = 3 x² - 5 x + 2

So, p(x) = 0

3 x² - 5 x + 2 = 0

(3 x - 2) (x - 1) = 0

3 x – 2 = 0

3 x = 2

x = 2/3

x – 1 = 0

x = 1

p (2/3) =(3 (2/3) - 2) ((2/3) - 1)

= (2-2) (-1/3)

= 0 (-1/3)

= 0

p (1) =(3(1) - 2) (1 - 1)

= (3 - 2) (0)

= (1) (0)

= 0

Hence the zeroes of p(x) are 2/3 and 1

Thus, Sum of zeroes = 5/3 and the product of zeroes = 2/3

From the basic relationships, we get

The sum of the zeroes = -coefficient of x/coefficient of x²

= -(-5)/3

= 5/3

The product of the zeroes = constant term/coefficient of x²

= 2/3

Thus the basic relationship verified. equation from roots solution3 equation from roots solution3 