Math, asked by genius1947, 1 month ago

What is a quadratic equation? Give 6 examples and solve them also.​

Answers

Answered by BRAINLYxKIKI
27

Provided Question :

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\tt{\orange{What\:is\:a\: Quadratic\: Equation\:?}} \\ \tt{\orange{Give\:6\:example\:and\:solve\: them}}

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Required Answer :

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\hookrightarrow A  \sf{\green{ Quadratic\: Equation}} is a quadratic polynomial of "x" variable in the form of \boxed{\bf{\blue{ax² \:+\: bx \:+\: c \:=\: 0 }}} , Where a , b , c are real numbers and a  \neq 0

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6 Examples :

  • \sf{ 2x² \:+\: kx \:+\: 3 \:=\: 0}

  • \sf{ kx ( x - 2 ) \:+\: 6 \:=\: 0 }

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

  • \sf{ 2x² + 8x - k³ \:=\: 0 }

  • \sf{ ( k - 3 )x² + 6x + 9 \:=\: 0 }

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

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The Solutions :

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ㅤㅤㅤ1: \sf{ 2x² \:+\: kx \:+\: 3 \:=\: 0}

Here ,

  •  \boxed{\sf{ a \:=\: 2 }}

  • \boxed{\sf{ b \:=\: k }}

  •  \boxed{\sf{ c \:=\: 3 }}

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 \because The given eqⁿ have equal roots

ㅤㅤㅤ\therefore \: \boxed{\sf{\purple{ b² \:-\: 4ac\:=\:0}}}

ㅤㅤ➪  \sf{ ( k )² \:-\: 4 ( 2 ) ( 3 ) \:=\: 0 }

ㅤㅤ➪  \sf{ k² - 8 ( 3 ) \:=\: 0 }

ㅤㅤ➪  \sf{ k² - 24 \:=\: 0 }

ㅤㅤ➪ \sf{ k² \:=\: 0 + 24 }

ㅤㅤ➪ \sf{ k² \:=\: 24 }

ㅤㅤ➪  \sf{ k \:=\: \sqrt{24} }

•°• ㅤㅤkㅤ=ㅤ\sqrt{24}

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ㅤㅤㅤ2: \sf{ kx ( x - 2 ) + 6 \:=\: 0 }

ㅤ ㅤㅤ :  \sf{ kx² - 2kx + 6 \:=\: 0 }

Here ,

  • \boxed{\sf{ a \:=\: k }}

  • \boxed{\sf{ b \:=\: 2k }}

  • \boxed{\sf{ c \:=\: 6 }}

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 \because The given eqⁿ have equal roots

ㅤㅤㅤ \therefore \boxed{\sf{\purple{ b² \:-\: 4ac\:=\:0}}}

ㅤㅤ➪ \sf{ ( 2k )² - 4 ( k ) ( 6 ) \:=\: 0 }

ㅤㅤ➪  \sf{ 4k² - 24k \:=\: 0 }

ㅤㅤ➪ \sf{ 4k² \:=\: 0 + 24k }

ㅤㅤ➪ \sf{ 4k² \:=\: 24k }

ㅤㅤ➪ \sf{ 4k \times k \:=\: 24k }

ㅤㅤ➪ \sf{ k \:=\: \dfrac{24k}{4k} }

ㅤㅤ➪ \sf{ k \:=\: \dfrac{\cancel{24}\cancel{x}}{\cancel{4}\cancel{x}} }

ㅤㅤ➪  \sf{ k \:=\: 6 }

•°•ㅤㅤkㅤ=ㅤ6

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ㅤㅤㅤ3: \sf{ x² - ( k + 4 )x + 2k + 5 \:=\: 0 }

Here ,

  • \boxed{\sf{  a \:=\: 1}}

  • \boxed{\sf{ b \:=\: ( k + 4 ) }}

  • \boxed{\sf{ c \:=\: 2k + 5 }}

 \because The given eqⁿ have equal roots

ㅤㅤㅤ \therefore \boxed{\sf{\purple{ b² - 4ac \:=\: 0 }}}

ㅤㅤ➪  \sf{ \{ ( k + 4 ) \}² - 4 ( 1 ) ( 2k + 5 ) \:=\: 0}

ㅤㅤ➪  \sf{ ( k + 4 )² - 4 ( 2k + 5 ) \:=\: 0 }

ㅤㅤ➪  \sf{ k² + 8k + 16 - 8k - 20 \:=\: 0 }

ㅤㅤ➪  \sf{ k² + \cancel{8k} + 16 - \cancel{8k} - 20 \:=\: 0 }

ㅤㅤ➪ \sf{ k² - 4 = 0 }

ㅤㅤ➪ \sf{ k² \:=\: 0 + 4 }

ㅤㅤ➪ \sf{ k \:=\: \pm \sqrt{4}}

ㅤㅤ➪  \sf{ k \:=\: \pm 2 }

•°•ㅤㅤkㅤ=ㅤ2 , -2

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ㅤㅤㅤ4: \sf{ 2x² + 8x - k³ \:=\: 0 }

Here ,

  • \boxed{\sf{ a \:=\: 2 }}

  • \boxed{\sf{  b \:=\: 8 }}

  • \boxed{\sf{ c \:=\: k³ }}

 \because The given eqⁿ have equal roots

ㅤㅤㅤ \therefore \boxed{\sf{\purple{  b² - 4ac \:=\: 0 }}}

ㅤㅤ➪ \sf{ ( 8 )² - 4 ( 2 ) ( k³ ) \:=\: 0 }

ㅤㅤ➪ \sf{ 64 - 8k³ \:=\: 0 }

ㅤㅤ➪  \sf{ - 8k³ \:=\: 0 - 64 }

ㅤㅤ➪ \sf{ \cancel{-} 8k³ \:=\: \cancel{-} 64 }

ㅤㅤ➪ \sf{ k³ \:=\: \dfrac{64}{8} }

ㅤㅤ➪  \sf{ k³ \:=\: \dfrac{\bcancel{64}}{\bcancel{8}} }

ㅤㅤ➪ \sf{ k \:=\sqrt[3]{8} }

•°•ㅤㅤkㅤ=ㅤ \pm 2

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ㅤㅤㅤ5: \sf{ ( k - 3 )x² + 6x + 9 \:=\: 0 }

Here ,

  • \boxed{\sf{ a \:=\: ( k - 3 ) }}

  •  \boxed{\sf{ b \:=\: 6 }}

  • \boxed{\sf{ c \:=\: 9 }}

 \because The given eqⁿ have equal roots

ㅤㅤㅤ \therefore \boxed{\sf{\purple{  b² - 4ac \:=\: 0 }}}

ㅤㅤ➪ \sf{ ( 6 )² - 4 \{ ( k - 3 ) \} ( 9 ) \:=\: 0 }

ㅤㅤ➪  \sf{ 36 -  4k + 12 ( 9 ) \:=\: 0 }

ㅤㅤ➪ \sf{ 36 - 4k + 108 \:=\: 0 }

ㅤㅤ➪  \sf{ - 4k + 144 \:=\: 0 }

ㅤㅤ➪ \sf{ - 4k \:=\: 0 - 144 }

ㅤㅤ➪  \sf{ \cancel{-} 4k \:=\: \cancel{-} 144 }

ㅤㅤ➪ \sf{ k \:=\: \cancel{\dfrac{144}{4}} }

ㅤㅤ➪ \sf{ k \:=\: 36 }

•°•ㅤㅤkㅤ=ㅤ36

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ㅤㅤㅤ6: \sf{ ( k - 12 )x² + 2(k - 12)x + 2 \:=\: 0 }

Here ,

  • \boxed{\sf{ a \:=\: ( k - 12 ) }}

  • \boxed{\sf{b \:=\: 2(k - 12)  }}

  • \boxed{\sf{ c \:=\: 2  }}

 \because The given eqⁿ have equal roots

ㅤㅤㅤ \therefore \boxed{\sf{\purple{ b² - 4ac \:=\: 0 }}}

ㅤㅤ➪  \sf{ \{ 2 ( k - 12 ) \}² - 4 ( k - 12 )( 2 ) \:=\: 0 }

ㅤㅤ➪ \sf{ 4 ( k - 12 )² - 8 ( k - 12 ) \:=\: 0 }

ㅤㅤ➪ \sf{ 4 ( k - 12 ) ( k - 12 ) ( k - 12 - 2 ) \:=\: 0 }

ㅤㅤ➪ \sf{ 4 ( k - 12 ) ( k - 14 ) \:=\: 0 }

ㅤㅤ➪  \sf{ ( k - 12 ) ( k - 14 ) \:=\: \dfrac{\cancel{0}}{\cancel{4}}}

ㅤㅤ➪ \sf{ ( k - 12 ) ( k - 14 ) \:=\: 0 }

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

ㅤㅤㅤㅤ➪ kㅤㅤ= 12ㅤorㅤ➪ kㅤㅤ= 14

Request* Sometime my LaTeX doesn't work properly , so please try to instruct me via Ratings ☞︎︎︎☜︎︎︎ // Happy Learning

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