find the zeros of p ( x ) = 2x^2-x-6 and verify the relationship of zeroes with these coefficients
Answers
Answer: is -3/2 and 2
Step-by-step explanation:
heyy my friend
2x square-x -6
by splitting the middle term,
2x square-4x+3x-6
2x(x-2) + 3(x-2)
(2x+3) (x-2) = 0
2x+3=0 & x-2=0
x= -3/2 and x=2
verification,
@+B= -b/a =1/2
@B=c/a=-3
and here @= alpha and B= beeta
<3
byee take care #staysafe
Given:
p ( x ) =2x² -x -6
To Find:
the zeros of p ( x ) = 2x² -x -6 and verify the relationship of zeroes with these coefficients
Solution:
2x² -x -6= 0
2x²-4x+3x -6 = 0
2x(x-2) + 3(x-2) =0
(x-2)(2x+3) =0 ( Taking x-2 common)
Now,
(x-2) =0
x=2
(2x+3) = 0
= -3/2
So, the two roots of the equation are 2 and -3/2
Now we will verify the zeroes and the polynomial
As we know,
Sum of zeroes of a polynomial = -b/a
Product of zeroes of a polynomial= c/a
where a = coefficient of x²
b = coefficient of x
c = constant term
Sum of zeroes = -b/a
2+(-3/2) = -(-1)/2
2+(-3/2) = 1/2
1/2 = 1/2
L.H.S = R.H.S
Product of zeroes = c/a
2(-3/2) = -6/2
-3= -3
L.H.S = R.H.S
Hence the zeroes of the polynomial are 2 and -3/2. The relationship between the zeroes and the polynomials is verified.