Find the value of k if one of the zeroes of the quadratic polynomial

(k-1)x2 + k x + 1 is -3.​

2 thoughts on “Find the value of k if one of the zeroes of the quadratic polynomial<br /><br /> (k-1)x2 + k x + 1 is -3.​”

1. SOLUTION

   TO DETERMINE

   The value of k if one of the zeroes of the quadratic polynomial (k-1)x² + kx + 1 is – 3.

   EVALUATION

   Here the given polynomial is

   (k-1)x² + kx + 1

   Now it is given that one of its roots is – 3

   So by the given condition

   [tex] \sf{(k – 1) {( – 3)}^{2} + k \times ( – 3) + 1 = 0}[/tex]

   [tex] \sf{ \implies \: 9(k – 1) – 3 k + 1 = 0}[/tex]

   [tex] \sf{ \implies \: 9k – 9 – 3 k + 1 = 0}[/tex]

   [tex] \sf{ \implies \: 6k – 8 = 0}[/tex]

   [tex] \sf{ \implies \: 6k = 8 }[/tex]

   [tex] \displaystyle \sf{ \implies \: k = \frac{8}{6} }[/tex]

   [tex] \displaystyle \sf{ \implies \: k = \frac{4}{3} }[/tex]

   FINAL ANSWER

   [tex] \displaystyle \sf{ \: k = \frac{4}{3} }[/tex]

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2. Given : one of the zeroes of the quadratic polynomial ( k-1)x² + k x + 1 is -3.​

   To Find : Value of k

   Solution:

   one of the zeroes of the quadratic polynomial ( k-1)x² + k x + 1 is -3.​

   Let say P(x) = ( k-1)x² + k x + 1

   -3 is one of the zero

   => P(-3) = 0

   => ( k-1)(-3)² + k (-3) + 1 = 0

   => 9 k – 9 – 3k + 1 = 0

   => 6k = 8

   => k = 8/6

   => k = 4/3

   Value of k is 4/3

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