Question: If one zero of the quadratic polynomial 2x^2 - 6kx +6x -7 is negative of the other then k =

Posted by: on 29.04.2021

Since, one root is negative of the other, here the sum of those two roots = 0

As per the given polynomial, the sum of the two roots = (-b)/a = [6(1-k)]/2 = 0 .

Or, 3.(1 - k) = 0 .   => 1 - k = 0 .

Hence, k = 1   .   (Ans.)

2x²-6kx+6x-7=0 Let the two zeroes be A and -A as given in the question Putting A in the equation 2A²-6kA+6A-7=0 -------- (i) Putting -A in the equation 2(-A)² - 6k(-A)+6(-A) -7=0 2A² - 6kA - 6A - 7=0 --------(ii) On subtraction i and ii 2A² - 6 kA + 6A -7=0 -2A² - 6kA + 6A +7=0 We get -12kA +12 A = 0 Hence k=1
.

In this question both the zeros are of opposite sign, Let z1 (zero 1) = z and z2 = -z

P: 2x^2 - 6kx  +6x -7

P(z) : 2z^2 - 6Kz +6z -7 = 0

P(-z): 2z^2 +6Kz - 6z - 7 =0

subtracting both the equation

(2z^2 and -7 being of same sign and value in both the equations cance out)

Leaving (-6Kz -6Kz)+ (6z+6z) = 0

-12kz +12z =0

-12kz = -12z

K =1

K=1

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