Question

#Quality Question
@Complex Numbers and Quadratic Equations
$ Class ----} 11th

Find number such that their sum is 6 and product is 14.
​

Answer 1

▪Assumption :-

Let the two numbers be α and β.

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▪Given :-

α + β = 6

and

αβ = 14.

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▪Solution :-

We have,

$\sf \alpha + \beta = 6 \\ \\ \implies \bf \beta = 6 - \alpha \: \: \: \:. \: .\: . \: \{i \}$

Also,

$\bf \alpha \beta = 14 \: \: \: \: . \: . \: . \: \{ii \}$

Putting {i} in {ii} We get,

$\alpha (6 - \alpha ) = 14 \\ \\ \implies 6\alpha - { \alpha }^{2} = 14 \\ \\ \implies { \alpha }^{2} - 6 \alpha + 14 = 0$

Which is a Quadratic Equations in α.

Using Quadratic Formula,

$\alpha = \dfrac{ - ( - 6) \pm \sqrt{( - 6) {}^{2} - 4 \times 1 \times 14 } }{2} \\ \\ = \frac{6 \pm \sqrt{36 - 56} }{2} \\ \\ = \frac{6 \pm \sqrt{ - 20} }{2} \\ \\ = \frac{6 \pm \sqrt{20} \: i}{2} \\ \\ = \frac{6 \pm2 \sqrt{5} \: i }{2}$

$\Large\implies \purple{\underline{ \boxed{ \bold{ \alpha = 3 \pm4 \mathcal{i}}}}}$

Putting Value of α in equation {i}

$\text{When} \: \: \alpha = 3 + \sqrt{5} \: i \\ \\ \beta = 3- \sqrt{5} \: i$

$\text{When} α = 3 - \sqrt{5} \: i \\ \\ \beta = 3 + \sqrt{5} \: i$

Hence, Two numbers are :-

$3 + \sqrt{5} \: i \: \: \bold{and} \: \: 3 - \sqrt{5} \: i$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Given that , The sum of number is 6 & the Product of number is 14 .

Need To Find : The two numbers ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Consider two numbers be α and β

As , We know that,

• The sum of numbers is 6 .

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Sum \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha + \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\} \:\: \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\}\:\:=\:\:\:\: 6 \\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \:\beta \: \:\:=\:\:6 \: - \alpha \:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

• The Product of numbers is 14 .

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Product \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha \beta \;)\:} \:: \: }}}$

$\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\} \:\: \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\}\:\:=\:\:\:\:14$

$\dashrightarrow \: \alpha \:\:\beta \: \:\:=\:\:14 \:\: \\\\ \qquad \underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Value \:of \:\alpha \::}}$

$\dashrightarrow \: \alpha \:\:\:( \: 6 - \alpha \:) \: \:\:=\:\:14 \:\:$

$\dashrightarrow \: 6\alpha \:- \alpha^2 \: \: \:\:=\:\:14 \:\: \\\\ \dashrightarrow \: 6\alpha \:- \alpha^2 \: \:- 14 \:\:=\:\:0 \:\qquad\pmb{\bf Quadratic \:Polynomial }\:$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━

$\qquad \qquad \underline{\pmb{\mathbb{\bigstar \:\:QUADRATIC \:\:POLYNOMIAL \:\::\:} \:\sf \alpha^2 - 6 \alpha - 14 }}$

⠀⠀Where ,

• a = 1 ,
• b = -6 &,
• c = - 14 .

As , We know that ,

⠀⠀⠀━ Formula To Calculate zeroes of Quadratic Polynomial :

$\qquad \star \:\:\underline {\boxed {{\pmb{\sf{  \:\:QUADRATIC \:POLYNOMIAL \:\:= \:\dfrac{ \: -b \sqrt{b^2 - 4ac } }{2a } \:  \:\:\:}}}}}$

$\qquad:\implies \sf \dfrac{ \: -b \sqrt{b^2 - 4ac } }{2a } \: \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf \dfrac{ \: -b \sqrt{b^2 - 4ac } }{2a } \: \:$

$\qquad:\implies \sf \dfrac{ \: -(-6) \sqrt{(-6)^2 - 4(1)(-14) } }{2(1) } \: \:$

$\qquad:\implies \sf \dfrac{ \: 6 \sqrt{36 - 4(-14) } }{2 } \: \:$

$\qquad:\implies \sf \dfrac{ \: 6 \sqrt{36 - 56 } }{2 } \: \:$

$\qquad:\implies \sf \dfrac{ \: 6 \sqrt{-20 } }{2 } \: \:$

$\qquad:\implies \sf \dfrac{ \: 6 \pm \sqrt{20 } }{2 } i \: \:$

$\qquad:\implies \sf \dfrac{ \: 6 \pm 2\sqrt{5 } }{2 } i \: \:$

$: \implies \underline {\boxed {\pmb{\pink{ \frak { \: \alpha =\:\:\:3 + \:\ \:\:\sqrt{5}i \:}}}}}\:\:\bigstar$

As , We know that,

When ,

$\dashrightarrow \: \:\alpha \: \:\:=\:\:\:3 + \:\ \:\sqrt{5}i \:$

Then ,

$\dashrightarrow \: \:\beta \: \:\:=\:\:\:3 - \:\ \:\:\sqrt{5}i \:$

$\qquad \therefore \underline {\sf Hence, \:Two \:numbers \:are \: \pmb{\bf \:3 + \:\ \:\:\sqrt{5}i \:\& \:\:3 \:- \:\:\sqrt{5}i\:}\:, \:respectively \:}$