Question

Find the zeroes of polynomial
${a }^{2} + 7a + 10$
and verify the relationship between the zeroes and the coefficients.

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Answer 1

Solution :-

a² + 7a + 10

To find zeroes equate a² + 7a + 10 to 0

a² + 7a + 10 = 0

a² + 5a + 2a + 10 = 0

a(a + 5) + 2(a + 5) = 0

(a + 5)(a + 2) = 0

a + 5 = 0, a + 2 = 0

a = - 5, a = - 2

Therefore - 5 and - 2 are zeroes of the polynomial

So

• α = - 5

• β = - 2

i) Sum of zeroes = α + β = - 5 + (-2) = - 5 - 2 = - 7

- b/a = - Coefficient of a/Coefficient of a² = - 7/1 = - 7

Therefore α + β = - b/a

ii) Product of zeroes = αβ = - 5(-2) = 10

c/a = Constant term/Coefficient of a² = 10/1 = 10

Therefore αβ = c/a

Hence relation is verified.

Answer 2

Answer:

Step-by-step explanation:

Given ,

a² + 7a + 10

The zero of the polynomial ,p(x)

p(x) = 0

a² + 7a + 10 = 0

Factorising,

a² + 2a+5a+10 = 0

a(a+2) + 5(a+2) = 0

(a+2)(a+5)=0

a+2=0 or a+5=0

a=-2 or a = -5

Let α= -2 and β= -5

Sum of zeroes , α+β= -2+(-5) = -7

Product of zeroes , αβ = -2*-5 = -10

Comparing p(a)= a² + 7a + 10 with the general form of polynomial,p(a) = ax^2+bx+c , we have

a = 1 , b = 7 , c = 10

Sum of zeroes ,α+B = -coefficent of x/coefficent of a^2

=> -b/a

=> -7/1

=> 7

Product of zeroes ,αβ = Constant /coefficient of a^2

=> c/a

=> 10/1

=> 10

Hence the relationship is verified