Question

Hey Guyz!

Here's the question :-

✯ If the zeroes of the quadratic polynomial p(x) = 3x² + (2k - 1)x - 5 are equal in magnitude but opposite in signs then find the value of K.

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Answer 1

ANSWER

$k = \frac{1}{2}$

METHOD:

in the attachment

IDENTITIES USED:

$> sum \: of \: roots \: = \frac{ - b}{a} = \frac - {coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

Answer 2

$\boxed{\textbf{\large{Step-by-step explanation}}}$

◾As we have given condition, the zeroes of the quadratic polynomial p(x) = 3x² + (2k - 1)x - 5 are equal in magnitude but opposite in sign

So, let's consider the zeros are α and β

◾compare the polynomial [3x² + (2k - 1)x - 5] with the standard form[ ax^2 + bx+c = 0 ]

Therefor ,

a = 3

b = ( 2k - 1 )

c = - 5

◾we know the relation between coefficients and roots of the equation is,

α + β = -b / a

and, α x β = c / a

so, α x β = c / a = (-(-5 )/ 3))

α x β = 5/ 3 ..............(1)

◾As we have given, the zeros of the polynomial are equal in magnitude but opposite in sign , so from this,

α = - β

put this in equation (1)

α x β = 5/ 3

(-β) x (β ) = 5 /3

- β^2 = 5/ 3

- β^2 = 5/3

β = - √(5/3)

◾put this value of root in a given polynomial

p(x) = 3x² + (2k - 1)x - 5

p ( -√(5 /3 ))

= 3(-√(5/3))^2 + (2k - 1 )(-√(5/3)) - 5

◾equate the equation with zero

Therefor,

⟹ 3(-√(5/3))^2 + (2k - 1 )(-(√(5/3)) ) - 5 = 0

⟹ 3 ( 5 / 3 ) + ( 2k - 1 )(-(√( 5 /3))) - 5 = 0

⟹ 5 + ( 2 K - 1 ) ( -(√(5 / 3) ) - 5 = 0

⟹ ( 2k - 1 ) ((-√ (5 / 3) ) = 0

⟹ 2k ((-√( 5 / 3 ))) - ( -√(5 / 3) ) = 0

⟹ 2k (-√( 5 / 3) ) = ( -√(5 / 3) )

⟹ 2k = [(-√( 5 /3 ) / -(√( 5 / 3) )]

⟹ 2k = 1

$\boxed{\textbf{\large{K = 1 / 2}}}$

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$\boxed{\textbf{\large{verification}}}$

◾Substitute the values of β = x = - √(5/3) and K = 1/ 2 in the given polynomial and equate the polynomial with zero .

Therefor,

⟹ 3x² + (2k - 1)x - 5 = 0

⟹ 3 ( - √(5/3) )^2 + ( 2 ( 1/ 2) - 1 )( - √(5/3) ) - 5 = 0

⟹ 3 ( 5 / 3 ) + ( 1 - 1 ) ( -√( 5 / 3)) - 5 = 0

⟹ 5 + ( 0 ) - 5 = 0

⟹ 0 = 0

hence verified

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