If a,b are the zeroes of polynomial f(x) = x²+5x+8 ,then a+b is

1.	If a,b are the zeroes of polynomial f(x) = x²+5x+8 ,then a+b is
Answer» Answer : The value of (a + b) is -5 Given : 'a' and 'b' are the zeroes if the polynomial x² + 5x + 8 To Find : The value of (a + b) CONCEPT to be USED : Relationship between the zeroes and the COEFFICIENTS of the polynomial $\sf \bullet \: \: Sum \: of \: the \: zeroes = -\dfrac{Coefficient \: of \: x}{Coefficient \: of \: x^{2}} \\\\ \sf \bullet \: \: Product \: of \: the \: zeroes = \dfrac{Constant \: term}{Coefficient \: of \: x^{2}}$ Solution : The given quadratic equation is $\sf x^{2} + 5x + 8$ From the relation of sum of the zeroes we have : $\sf \implies a + b = -\dfrac{5}{1} \\\\ \sf \implies a + b = -5$ Thus , the value of (a + b) is -5

Answer»

Answer :

The value of (a + b) is -5

Given :

'a' and 'b' are the zeroes if the polynomial x² + 5x + 8

To Find :

The value of (a + b)

CONCEPT to be USED :

Relationship between the zeroes and the COEFFICIENTS of the polynomial

$\sf \bullet \: \: Sum \: of \: the \: zeroes = -\dfrac{Coefficient \: of \: x}{Coefficient \: of \: x^{2}} \\\\ \sf \bullet \: \: Product \: of \: the \: zeroes = \dfrac{Constant \: term}{Coefficient \: of \: x^{2}}$

Solution :

The given quadratic equation is

$\sf x^{2} + 5x + 8$

From the relation of sum of the zeroes we have :

$\sf \implies a + b = -\dfrac{5}{1} \\\\ \sf \implies a + b = -5$

Thus , the value of (a + b) is -5

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