find the zeroes of the quadratic polynominal xsq.+7x+10 and verify the relationship between the zeroes what the coeffients
Answers
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:
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. ✌