Question

find the zeroes of the quadratic polynominal xsq.+7x+10 and verify the relationship between the zeroes what the coeffients
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Answer 1

Given :

• x² + 7x + 10

To find :

• zeroes of the Qudratic polynomial

Solution:

Given , x² + 7x + 10

• Let p(x) = x² + 7x + 10

➻ p(x) = x² + 7x + 10

➻ p(0) = x² + 7x + 10

putting the value

➻ x² + 7x + 10 = 0

Now splitting middle term

➻ x² + 2x + 5x + 10 = 0

➻ x(x + 2) + 5(x +2) = 0

➻ (x + 2) (x + 5) = 0

➻ x = -2 or x = -5

so , α is -2 , β is -5 are zeroes of polynomial

Now,

➻ p(x) = x² + 7x + 10

➻ 1x² + 7x + 10

Now comparing the Qudratic equation ax² + bx + c so, a is 1 , b is 7x , c is 10

Now we have to verify the sum of zeroes

• sum of zeroes = co effiecent of x/ co effiecent of x²
• α + β = b/a

α + β = b/a

➻ LHS = α + β

= -2 + (-5)

= -2 - 5

= -7

➻ RHS = -b/a

= - 7/1

= -7

Now we have to verify the product of zeroes

• product of zeroes = constant term / co effiecent of x²
• α × β = c/a

α × β = c/a

➻RHS = α × β

= (-2) (-5)

= 10

➻ RHS = c/a

= 10/1

= 10

Hence, verified LHS = RHS

Answer 2

Answer:

x²+7x+10=0

x²+2x+5x+10=0

x(x+2)+5(x+2)=0

(x+2)(x+5) = 0

x+2 = 0 ; x = -2

x+5 = 0 ; x = -5

Relationship between the zeroes and coefficients :-

Sum of zeroes = -2+(-5) = -2-5

= -7/1 = -x coefficient /x² coefficient

Product of zeroes = (-2)(-5)

= 10/1 = constant/x² coefficient

Step-by-step explanation:

Hope this answer will help you. ✌