Question

if one of the zeros of the quadratic polynomial (k-1)x^2+kx+1 is -3 then value of k is A)4/3 B)-4/3 C)2/3 D)-2/3 EXPLAIN

Answer 1

Gɪᴠᴇɴ :-

• one of the zeros of the quadratic polynomial (k-1)x^2+kx+1 is (-3) .

Tᴏ Fɪɴᴅ :-

• Value of k ?

ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-

• If (x - a) is indeed a factor of p(x), then the remainder after division by (x - a) will be zero or f(a) will be Equal to Zero.

Sᴏʟᴜᴛɪᴏɴ :-

→ f(x) = (k-1)x^2+kx+1

Putting x = (-3) , we get,

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

→ (k - 1)9 - 3k + 1 = 0

→ 9k - 9 - 3k + 1 = 0

→ 9k - 3k - 9 + 1 = 0

→ 6k - 8 = 0

→ 6k = 8

→ k = (8/6)

→ k = (4/3) (Ans.) (Option A).

Hence, value of k will be (4/3).

Answer 2

✯✯ QUESTION ✯✯

if one of the zeros of the quadratic polynomial (k-1)x²+kx+1 is -3 then value of k is : -

A)4/3

B)-4/3

C)2/3

D)-2/3

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✰✰ ANSWER ✰✰

$\implies\tt{(k-1){x}^{2}+kx+1=0}$

➮Putting x = -3

$\implies\tt{(k-1){(-3)}^{2}+k(-3)+1=0}$

$\implies\tt{(k-1)9-3k+1=0}$

$\implies\tt{9k-9-3k+1=0}$

$\implies\tt{9k-3k=9-1}$

$\implies\tt{6k=8}$

$\implies\tt{k=\cancel\dfrac{8}{6}}$

$\implies\tt{\large{\boxed{\bold{\bold{\green{\sf{k=\dfrac{4}{3}}}}}}}}$

➥So,The value of k is 4/3 .

☛Option A)4/3 is Correct...