Question

alpha beta are the zeros of the polynomial x square + 5 x + 3 and alpha minus beta is equal to 3 then c equal to ​

Answer 1

Solution :-

Given that, α and β are the zeroes of the polynomial x^2 + 5x + c.

We know, a^2x + bx + c = 0.

By comparing both we get :-

a = 1

b = 5

c = c

Given,

$: \implies \sf \alpha - \beta = 3 \: \: \: \: \: \: \: \: \: \: \: \: \: ...(1)$

Sum of zeroes :-

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

$: \implies \sf \alpha + \beta = - \dfrac{5}{1}$

$: \implies \sf \alpha + \beta = - 5 \: \: \: \: \: \: \: \: \: \: \: \: \: ...(2)$

Product of zeroes :-

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

$: \implies \sf \alpha \times \beta = c\: \: \: \: \: \: \: \: \: \: \: \: \: ...(3)$

By adding eq(1) and eq(2) we get :-

$: \implies \sf 2 \alpha = - 2$

$: \implies \sf \alpha = - 1$

By substituting α = -1 in eq(1) we get :-

$: \implies \sf - 1 + \beta = - 5$

$: \implies \sf \beta = - 4$

Now substitute the value of α and β in eq(3) :-

$: \implies \sf - 1 \times - 4 = c$

$: \implies \bf c = 4$

Answer :- Option C

Answer 2

Given:

polynomial:

$x^2+5x+3$

$\alpha -\beta =3$

To find:

the value of c.

Solution:

General form of polynomials is:

$ax^2+bx+c$

In the given polynomial,

a=1, b=5

Now we know that,

Sum of the roots of a polynomial is the ratio of neagtive of b to a and the product of the two roots is the ratio of c to a. So we can write,

$\alpha+ \beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

Putting the values, we get,

$\alpha+ \beta =-\frac{5}{1}$

⇒$\alpha +\beta =-5$ -(1)

Given that,

$\alpha -\beta =3$ -(2)

Adding the two acquired equations (1) and (2), we get,

$2\alpha =-2$

⇒$\alpha =-1$

So,

$\alpha -\beta =3$

⇒$\beta =\alpha -3$

⇒$\beta =-1-4$

⇒$\beta =-4$

Now,

$\alpha \beta =\frac{c}{a}$

⇒$(-1)(-4)=\frac{c}{1}$

⇒$c=4$

Hence, the required value of c is 4.