Question

find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. x^2-9​

Answer 1

Answer:

x = ± 3

Note:

★ The possible values of the variable for which the polynomial becomes zero are called its zeros .

★ In order to find the zeros of the given polynomial , equate it to zero .

★ A quadratic polynomial can have atmost two zeros.

★ The general form of a quadratic polynomial is given by : ax² + bx + c .

★ If A and B are the zeros of the quadratic polynomial ax² + bx + c , then ;

• Sum of zeros , (A + B) = -b/a

• Product of zeros , (A•B) = c/a

Solution:

Here,

The given quadratic polynomial is ;

x² - 9

The given quadratic polynomial can be rewritten as ; x² + 0•x - 9

Clearly,

a = 1

b = 0

c = -9

Now,

Let's find the zeros of the given quadratic polynomial by equating it to zero .

Thus,

=> x² - 9 = 0

=> x² = 9

=> x = √9

=> x = ± 3

Now,

Sum of zeros = - 3 + 3 = 0

Also,

-b/a = -0/1 = 0

Clearly,

Sum of zeros = -b/a

Now,

Product of zeros = -3×3 = -9

Also,

c/a = -9/1 = -9

Clearly,

Product of zeros = c/a

Hence verified

Answer 2

ANSWER:

$P(x) = x^{2} - 9$

Therefore

a = 1

b = 0

c = -9

=> P(x) = 0

=> x² = 9

=> x = √9

=> x = ± 3

(Sum of zeros ) = $\alpha + \beta$ = $\frac{-b}{a}$

 $\alpha + \beta$ = - 3 + 3 = 0

$\frac{-b}{a} = \frac{-0}{1}$ = 0

(Product of zeros) = $\alpha \beta$ = $\frac{c}{a}$

$\alpha \beta$ = -3×3 = -9

$\frac{c}{a} = \frac{-9}{1}$= -9