Question

In the sum of the zeros of the
quadratic polynomid 3x2 - kx+ 6 is
3, then find the value of K.​

Answer 1

Given:-

• The zeroes of the polynomial p(x) = 3x² - kx + 6 are α and β.
• Sum of zeroes = α + β = 3

To find:-

• Value of k

Answer:-

▪In a polynomial ax² + bx + c having zeroes α and β, (α + β) = - (coefficient of x) / (coefficient of x²) = -b/a

▪ Given that, p(x) = 3x² - kx + 6, it can be written as,

p(x) = 3x² - kx + 6

→ p(x) = 3x² + (-k)x + 6

▪ We have, p(x) = 3x² + (-k)x + 6. On comparing this with the general form, i.e., p(x) = ax² + bx + c,

we get,

⦾ a = 3

⦾ b = -k

⦾ c = 6

▪ So, putting in the above formula, sum of zeroes = α + β = -b/a =

→ α + β = - (-k)/(3)

→ α + β = k/3 ----( 1 )

▪ Now, it is given that, sum of zeroes = α + β = 3

But from ( 1 ), α + β = k/3.

So, we can say that,

α + β = 3 = k/3

→ k = 9 Ans.

Other formulae:-

▪In a quadratic polynomial, ax² + bx + c, having roots α and β,

⦾ αβ = c/a

▪In a cubic polynomial, ax³ + bx² + cx + d, having roots α, β and γ,

⦾ α + β + γ = -b/a

⦾ αβ + βγ + γα = c/a

⦾ αβγ = -d/a

Answer 2

Answer:

${\underline{\boxed{ \sf{value \: of \: k= 9}}}}$

refer to the attachment for the explanation.

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MoRE information!!

❄️For Quadratic Polynomial:

If α and β are the roots of a quadratic polynomial ax2+bx+c, then,

▪️α + β = -b/a

▪️Sum of zeroes = -coefficient of x /coefficient of x²

▪️αβ = c/a

▪️Product of zeroes = constant term / coefficient of x²

❄️For Cubic Polynomial

If α,β and γ are the roots of a cubic polynomial ax3+bx2+cx+d, then

▪️α+β+γ = -b/a

▪️αβ +βγ +γα = c/a

▪️αβγ = -d/a

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