Question

Q2. Find the value of ‘k’ for which (3−k)x2 + 7 is a constant polynomial.

Answer 1

Step-by-step explanation:

Given x

2

−2x(1+3k)+7(3+2k)=0 has equal roots

As we know that

For the quadratic equation to have equal roots discriminant should be zero

⟹(−2(1+3k))

2

−4(1)(7(3+2k))=0

⟹4(1+3k)

2

−28(3+2k)=0

⟹(9k

2

+6k+1)−21−14k=0

⟹9k

2

−8k−20=0

⟹9k

2

−18k+10k−20=0

⟹9k(k−2)+10(k−2)=0

⟹(9k+10)(k−2)=0

⟹k=−

9

10

,2

Answer 2

Step-by-step explanation:

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