Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients
49x^2 - 81

Answer 1

In order to find the zeroes of the given quadratic polynomial, we've to factorise :

$49 {x}^{2} - 81$

$49 \: can \: be \: written \: as \: {7}^{2} \\ \\ and \: 81 \: can \: be \: written \: as \: {9}^{2}$

Now,

$49 {x}^{2} - {9}^{2} \\ \\ (7x \: + 9)(7x - 9)$

FIND ZEROES :

7x + 9 = 0

=> 7x = - 9

•°• x = -9/7

Also,

7x - 9 = 0

=> 7x = 9

•°• 9/7

Now,

VERIFICATION :

Here,

• a = 1
• b = 0
• c = -81

•) Sum of zeroes = - b/a

=> -9/7 + 9/7 = - ( 0 )/49

=> 0 = 0

L.H.S = R.H.S

•) Product of zeroes = c/a

=> 9/7 × ( - 9 )/7 = - 81/49

=> -81/49 = -81/49

L.H.S = R.H.S

Hence Proved!

$\rule{100}2$