1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
(1) x2 – 2x - 8
(ii) 452 – 4s +1
(iii) 6x2-3-7x
Answers
Answer:
hope this helps u
Step-by-step explanation:
1)P(X) => X²-2X-8
=> X²-4X+2X-8
=> X(X-4) +2( X-4)
=> (X-4) (X+2) = 0
=> (X-4) = 0 OR (X+2) = 0
=> X = 4 OR X = -2
4 and -2 are the two zeroes of the polynomial X²-2X-8.
RELATIONSHIP BETWEEN THE ZEROES AND COEFFICIENT.
Sum of zeroes = Alpha + Beta = -Coefficient of X/Coefficient of X².
= 4 + (-2) = 4-2 = 2/1
2=2
And,
Product of zeroes = Alpha × Beta = = Constant term/Coefficient of X².
4 × -2 = -8/1
-8=-8
2)mate second question is 4s^2-4s+1
4s²-2s-2s+1
2s[2s-1] -1[2s-1]
2s-1 = 0,2s-1 = 0
s = 1/2,1/2
Sum of zeroes = Alpha + Beta = -Coefficient of X/Coefficient of X².
1/2+1/2 = -[-]4/ 4
2/2=4/4
1=1
Product of zeroes = Alpha × Beta = = Constant term/Coefficient of X²
1/2*1/2 = 1 /4
1/4=1/4
3)arrange it in correct manner that is degree is in descending order
given,
6x²-7x - 3
to find the zeroes of the polynomial,
6x²-7x - 3 = 0
6x²-9x +2x -3 = 0
3x(2x-3) + 1(2x -3) = 0
(3x+1) (2x-3) = 0
x = -1/3 or 3/2
sum of the zeroes = - ( co-efficient of x) / co-efficient of x²
= 3/2 - 1/3 = 7/6 lcm of 2,3 is 6
= 9/6-2/6 =7/6
=7/6=7/6
product of the zeroes = constant / co-efficient of x²
-1/3 × 3/2 = -3/6
-3/6=-3/6
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