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Answer Question No. 5th's Sub part

The
Question Is From Quadratic Equations

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Answer 1

Given:-

• 9x² - 24x + k = 0.
• 2kx² - 40x +25 = 0.

To Find:-

• The value of k for which the equⁿ has equal roots.

Solution :-

Firstly of a equⁿ in standard form of ax² + bx + c , we know that the roots are equal if the Discriminant of the equⁿ is equal to 0 i.e. b² - 4ac = 0.

➤ (i) 9x² - 24x + k = 0 .

With respect to Standard form ax² + bx + c = 0 ,

• a = 9
• b = (-24)
• c = k .

Now here b² - 4ac must be equal to 0.

⇒ b² - 4ac = 0.

⇒ (-24)²- 4 × 9 × k = 0

⇒ 576 - 36 k = 0

⇒ 576 = 36k

⇒ k = 576 / 36 .

⇒ k = 16.

Hence the value of k must be 16.

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➤ (ii) 2kx² - 40x + 25 = 0.

With respect to Standard form ,

• a = 2k .
• b = (-40)
• c = 25.

⇒ b² - 4ac = 0.

⇒ (-40)² - 4 × 2k × 25 = 0.

⇒ 1600 - 200k = 0.

⇒ 200k = 1600.

⇒ k = 1600/200

⇒ k = 8 .

Hence the value of k should be 8.

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Also have a look at this table.

$\boxed{\begin{tabular}{|c|c|} Conditions &Nature of Roots \\ \cline{1-2} D > 0 & Roots are real \\ \cline{1-2} D=0 & Roots are equal \\ \cline{1-2} D < 0 & Roots are complex nos. \end{tabular}}$