Question

5. Find the zeroes of the quadratic polynomial : x2 – 2x – 8 and verify the relationship between
the zeroes and the coefficients.​

Answer 1

Note:

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

★ A quadratic polynomial can have atmost two zeros .

★ The general form of a quadratic polynomial is given as ; ax² + bx + c .

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

• Sum of zeros , (α + ß) = -b/a

• Product of zeros , (αß) = c/a

Solution :

Here ,

The given quadratic polynomial is ;

x² - 2x - 8

Now ,

Comparing the given quadratic polynomial with the general quadratic polynomial ax² + bx + c , we have ;

a = 1

b = -2

c = -8

Now ,

• -b/a = -(-2)/1 = 2

• c/a = -8/1 = -8

Now ,

Let's find the zeros of the given qudratic polynomial .

=> x² - 2x - 8 = 0

=> x² - 4x + 2x - 8 = 0

=> x(x - 4) + 2(x - 4) = 0

=> (x - 4)(x + 2) = 0

=> x = 4 , -2

Now ,

• Sum of zeros = 4 + (-2) = 4 - 2 = 2 = -b/a

• Product of zeros = 4×(-2) = -8 = c/a

Clearly ,

Sum of zeros = -b/a and

Product of zeros = c/a

Hence verified .

Answer 2

Given:-

• $\sf f(x)$ = $\sf x²-2x-8$

To find:-

• Zeroes of this quadratic equation
• Verification of its zeros

Solution:-

$\implies$ $\sf x²-2x-8$

$\implies$ $\sf x²-(4-2)x-8$

$\implies$ $\sf x²-4x+2x-8$

$\implies$ $\sf x(x-4)+2(x-4)$

$\implies$ $\sf (x-4)(x+2)$

• hence, α = 4 and ẞ = -2

Verification of zeros:-

Sum of the zeroes:-

$\implies$ $\alpha + \beta =$ $\sf\dfrac{-b}{a}$

$\implies$ $\sf 4+(-2)$ = $\sf\dfrac{-(-2)}{1}$

$\implies$ $\sf 2$ = $\sf 2$

L.H.S = R.H.S

Product of zeros:-

$\implies$ $\alpha \times \beta$ = $\sf\dfrac{c}{a}$

$\implies$ $\sf 4(-2)$ = $\sf\dfrac{-8}{1}$

$\implies$ $\sf -8$ = $\sf -8$

L.H.S = R.H.S

hence, verified