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Questions
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients.
6x2 – 3 – 7x
Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients.
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Solution 1
6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3)(3x + 1)
= `2(x - 3/2)3(x+1/3)`
For p(x) = 0 we have,
Either (3x + 1) = 0
`x = -1/3`
or (2x – 3) = 0
`x = 3/2`
Thus, the zeroes of
6x2 – 3 – 7x are `-1/3 "and" 3/2`
⇒ Sum of the zeroes = `"-Coefficient of x"/("Coefficient of" x^2)`
⇒ `-1/3 + 3/2= (- (-7))/6`
⇒ `7/6 = 7/6`
Product of the zeroes = `"Constant term"/("Coefficient of "x^2)`
= `-1/3 xx 3/2=(-3)/6`
⇒ `-1/2 = -1/2`
Thus, the relationship between the zeroes and coefficients in the polynomial 6x2 – 3 – 7x is verified.
Solution 2
Given, quadratic polynomial:
p(x) = 6x2 – 3 – 7x
For zeroes of polynomial, put p(x) = 0
∴ 6x2 – 7x – 3 = 0
By splitting the middle term,
6x2 – 9x + 2x – 3 = 0
3x(2x – 3) + 1(2x – 3) = 0
(2x – 3) (3x + 1) = 0
∴ 2x – 3 = 0 and 3x + 1 = 0
or `x = 3/2` and `x = - 1/3`
Therefore, `α = 3/2` and `β = - 1/3` are the zeroes of the given polynomial.
Verification:
Sum of zeroes = α + β
= `3/2 + (-1/3)`
= `3/2 - 1/3`
= `(9 - 2)/6`
= `7/6`
= `-(("Coefficient of" x))/("Coefficient of" x^2)`
and Product of zeroes = αβ
= `(3/2)(-1/3)`
= `-1/2`
= `"Constant term"/("Coefficient of" x^2)`
Hence Verified.
Notes
Students should refer to the answer according to the question.
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