Question

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

Answer 1

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

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