if one of the zeroes of the quadratic polynomial (k-1) x2+1 is -3 then the value of k is
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Given that P(x) = ( k -1 ) x² + kx + 1 = 0.
→ p( -3 ) = ( k – 1 ) (-3)² + k (-3) + 1 = 0.
⇒ ( k – 1 ) × 9 -3 k + 1 = 0.
⇒ 9k – 9 – 3k + 1 = 0.
⇒ 6k – 8 = 0.
⇒ 6k = 8.
Therefore, k = 8/6 = 4/3
→ p( -3 ) = ( k – 1 ) (-3)² + k (-3) + 1 = 0.
⇒ ( k – 1 ) × 9 -3 k + 1 = 0.
⇒ 9k – 9 – 3k + 1 = 0.
⇒ 6k – 8 = 0.
⇒ 6k = 8.
Therefore, k = 8/6 = 4/3
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