Question

If the sum of the zeroes of the polynomial
p(x) = (a+1)x²+(2a+3)x+(3a+4) is –1. Then find the product of its zeroes.

Answer 1

Answer:

Sum of zeroes = (-1)

We know that:

Sum of zeroes = -b/a

.°. (-1) = (-2a - 3)/(a + 1)

=> - a - 1 = - 2a - 3

=> - a + 2a = 1 - 3

=> a = -2

Product of zeroes = c/a

.°. Product = (3a + 4)/(a + 1)

Put a = -2:

=> Product = (3(-2) + 4)/(-2 + 1)

=> Product = (-6 + 4)/(-1)

=> Product = -2/(-1)

=> Product = 2

Answer 2

$\huge \fcolorbox{black}{lightblue}{Solution.}$

Here a = a + 1, b = 2a + 3, c = 3a + 4

Sum of zeroes = –1

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

b = a

2a + 3 = a + 1

2a – a = 1 – 3

a = – 2

∴ Product of zeroes

$\frac{c}{a} = \frac{3a + 4}{a + 1}$

Putting a = – 2,

$\frac{3( - 2) + 4}{ - 2 + 1} = \frac {- 6 + 4}{ - 1}$

$\frac{ - 2}{ - 1} = 2$