Question

if a and bare thezero of the polynomial 2y^2+7y+5 write the value of a+b+ab​

Answer 1

Given: The x and b are zeroes of

the polynomial

$2 {y}^{2} + 7y + 5$

Answer:

$2 {y}^{2} + 7y + 5 = 0 \\ {y}^{2} + 7y + 10 = 0 \\ {y}^{2} + 5y + 2y + 10 = 0 \\ y(y + 5) + 2(y + 5) = 0 \\ (y + 2)(y + 5) = 0 \\ therefore \: the \: zero \: \alpha = 2 \: and \: \beta = 5$

$then \: \alpha + \beta + \alpha \times \beta = \\( 2 + 5) + (2 \times 5) \\ 7 + 10 = 17 \\ i \: hope \: this \: will \: help \: you \: then \: \\ pls \: follow \: me \: and \: mark \: it \: as \\ \bold \blue {\mathfrak{brainliest}}$

Answer 2

Given:

We have been given that a and b are the zeroes of the polynomial 2y^2 + 7y + 5.

To Find:

We need to find the value of a + b + ab.

Solution:

We have been given a polynomial 2y^2 + 7y + 5.

Here, a = 2, b = 7 and c = 5.

As it is given that a and b are two zeroes,

So, sum of zeroes(a + b)

= -b/a

= -7/2___(1)

Product of zeroes(ab)

= c/a

= 5/2___(2)

Now we need to find the value of a + b + ab,

So we have,

a + b + ab = (a + b) + ab

Substituting the values from equation 1 and 2, we get,

(-7/2) + (5/2)

= (-7 + 5)/2

= -2/2

= -1

Therefore, the value of a + b + ab is -1.