Question

If the roots of 2z^2-6x+k =0 are real and equal the find k​

Answer 1

$\Huge \bf \green{Solution:-}$

The roots of the quadratic equation are real and equal.

$\bf ∴ b² - 4ac = 0$

$\bf (-6)² - 4 × 2 × k = 0$

$\bf - 8k = -36$

$\bf k = \dfrac{9}{ 2}$

Answer 2

$\LARGE\mathfrak{\underline{\underline{✪\: \: \: Answer :-}}}$

$\large\: \bigstar\textsf\textcolor{orange}{\: \: \: 2z² - 6x + k = 0}$

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➛ Comparing this equation with :-

$\large\: \bigstar\textsf\textcolor{orange}{\: \: \: az² + bz + c = 0 }$

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➛ Here :-

$\large: \: \Longrightarrow\textsf{\: \: \: a = 2 \: \: \: b = - 6 \: \: \: c = k}\\\\\\\large\: \bigstar\textsf\textcolor{orange}{\: \: \: ∆ = b² - 4ac . }\\\\\\\large: \: \Longrightarrow\textsf{= ( - 6 )² - 4 × 2 × k}\\\\\\\large : \: \Longrightarrow\underline{\boxed{\textsf\textcolor{magenta}{∆ = 36 - 8k}}}$

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• The roots are real and equal so , ∆ must be ' zero ' .

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$\large: \: \Longrightarrow\textsf{- 8k + 36 = 0}\\\\\\\large: \: \Longrightarrow\textsf{- 8k = -36 }\\\\\\\large: \: \Longrightarrow\textsf{ k = \cfrac{- 36}{- 8} }\\\\\\\large: \: \Longrightarrow\textsf{ k = \cancel\cfrac{- 36 }{ - 8 } }\\\\\\\large \: \therefore\underline{\boxed{\textsf\textcolor{magenta}{ k =\cfrac { 9 }{2} }}}$

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