Question

Solve the following having equal roots :

1: x² - ( k + 4 )x + 2x + 5 = 0
2: ( k - 12 )x² + 2 ( k - 12 )x + 2 = 0

Spam = 20 answers report
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Answer 1

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1. $\sf{\red{x² - ( k + 4 )x + 2x + 5 \:=\: 0}}$

2. $\sf{\red{( k - 12 )x² + 2 ( k - 12 )x + 2 \:=\:0}}$

${\fcolorbox{blue}{black}{\green{ Here\:are\:the\:Answers\:: }}}$

1. $\sf{x² - ( k + 4 )x + 2x + 5 \:=\: 0}$

Here ,

ㅤa = 1ㅤ,ㅤb = -( k + 4 )ㅤ,ㅤc = ( 2x + 5 )

°•° The given equation have equal roots

•°•ㅤㅤㅤ$\boxed{\sf{\blue{b² \:-4ac\:=\:0}}}$

➪ {-( k + 4 )}² - 4 × ( 1 ) × ( 2x + 5 ) = 0

➪ k² + 8k + 16 - 8k - 20 = 0

➪ k² + 16 - 20 = 0ㅤㅤ[ 8k are cancelled ]

➪ k² - 4 = 0

➪ k² ㅤ = 0 + 4

➪ kㅤㅤ= $\pm \sqrt{4}$

➪ kㅤㅤ= $\pm \sf{2}$

★ k = 2 , -2

2. $\sf{( k - 12 )x² + 2 ( k - 12 )x + 2 \:=\:0}$

Here ,

ㅤa = ( k - 12 ) , b = 2( k - 12 ) , c = 2

°•° The given equation have equal roots

•°• ㅤㅤㅤ$\boxed{\sf{\blue{b²\:-4ac\:=\:0}}}$

➪ {2•( k - 12 )}² - 4 × ( k - 12 ) × ( 2 ) = 0

➪ 4 ( k - 12 )² - 8 ( k - 12 ) = 0

➪ 4( k - 12 )( k - 12 ) ( k - 12 - 2 ) = 0

➪ 4 ( k - 12 ) ( k - 14 ) = 0

➪ ( k - 12 ) ( k - 14 ) = $\dfrac{0}{4}$

➪ ( k - 12 ) ( k - 14 ) = 0

•°• Either ( k - 12 ) = 0 ; ( k - 14 ) = 0

ㅤㅤㅤㅤㅤ➪ k = 12 ㅤ ;ㅤ➪ k = 14

ㅤㅤㅤ꧁ ʙʀᴀɪɴʟʏ×ᴋɪᴋɪ ꧂

Answer 2

Answer:

$\huge \mathfrak \anser = \to$

see the attached picture..hope this helps you (: