Question

find the value of k such that the quadratic polynomial (k - 12)x2 -2(k - 12)x + 2 has equal roots

Answer 1

Required Answer:-

When a quadratic equation have equal roots, the discriminate (D) = 0. Here, Discriminate represents b² - 4ac for a quadratic equation of the form ax² + bx + c.

Comparing with ax² + bx + c:

• a = k - 12
• b = -2(k - 12)
• c = 2

Plugging in the approached equation:

➙ b² - 4ac = 0 ($\because$ equal roots)

➙ {-2(k - 12)}² - 4(k - 12)2 = 0

➙ 4(k - 12)² - 8(k - 12) = 0

➙ 4(k - 12){k - 12 - 2} = 0

(Here I have taken 4(k - 12) as common for easier calculation)

➙ 4(k - 12)(k - 14) = 0

➙ (k - 12)(k - 14) = 0

Equating to 0,

➙ k = 12,14

Hence:-

The required values of k are 12 or 14.

Answer 2

$\large\underline{\red{\sf \orange{\bigstar} Correct\;Quesyion}}$

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• find the value of k such that the quadratic polynomial (k - 12)x2 -2(k - 12)x + 2 has equal roots

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$\large\underline{\green{\sf \blue{\bigstar} Solution}}$

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• Comparing with ax² + bx + c:

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• a = k - 12

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• b = -2(k - 12)

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• c = 2

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$\pink{\sf{\star\;☯\; Plugging\; in\; the \;approached\; equation:}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• ➙ b² - 4ac = 0 ($\because$ equal roots)

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• ➙ {-2(k - 12)}² - 4(k - 12)2 = 0

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• ➙ 4(k - 12)² - 8(k - 12) = 0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• ➙ 4(k - 12){k - 12 - 2} = 0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• ➙ 4(k - 12)(k - 14) = 0

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• ➙ (k - 12)(k - 14) = 0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\purple{\sf{\star\;☯\; Equating\; to\; 0,}}$

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• ➙ k = 12,14

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Hence:-

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$\red{\sf{\star\;☯\; The\; required \;values \;of\; k \;are\; 12\; or \;14.}}$

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