Question

If one of the zeroes of the quadratic polynomial ( k - 1 ) x² + kx + 1 is -3 then the value of k is a) 4/3 b) -4/3 c) 2/3 d) -2/3
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Answer 1

$\red{\boxed{\sf{Given:}}}$

one zero of the quadratic polynomial [ ( k - 1 ) x² + kx + 1 ] is (-3).

$\red{\boxed{\sf{To\:\:be\:\:found:}}}$

The value of K.

P(x) = ( k - 1 ) x² + kx + 1

Now,

As (-3) is the zero of the quadratic polynomial P(x), then P(-3) = 0

→ P(-3) = ( k - 1 ) (-3)² + k(-3) + 1 = 0

→ ( k - 1 ) × 9 - 3k + 1 = 0

→ [ 9(k) - 9(1) ] - 3k + 1 = 0

→ 9k - 9 - 3k + 1 = 0

→ 9k - 3k - 9 + 1 = 0

→ 6k - 8 = 0

→ 6k = 8    [taking (-8) to RHS]

→ k =  $\bf \frac{8}{6}$

[dividing both numerator and denominator by 2 we get,]

$\boxed{\boxed{\bf \rightarrow k = \frac{4}{3}}}$

Hence,

Option a) 4/3 is the correct answer.

Answer 2

Answer:

✩ Polynomial : (k - 1)x² + kx + 1

One of Zero of Polynomial p(x) is - 3

⠀

☯ $\underline{\boldsymbol{According\: to \:the\: Question :}}$

$:\implies\sf p(x) = (k-1)x^2+kx+1\\\\\\:\implies\sf p(-\:3)=(k-1)x^2+kx+1=0\\\\\\:\implies\sf [k-1](-\:3)^2+k(-\:3)+1=0\\\\\\:\implies\sf [k-1]9 - 3k+1=0\\\\\\:\implies\sf 9k - 9 - 3k + 1 = 0\\\\\\:\implies\sf 6k - 8 = 0\\\\\\:\implies\sf 6k = 8\\\\\\:\implies\sf k = \dfrac{8}{6}\\\\\\:\implies\underline{\boxed{\sf k = \dfrac{4}{3}}}$

⠀

$\therefore\:\underline{\textsf{Hence, required value of k is a) \textbf{ {}^{\text4}\!/{}_{\text3} }}}.$