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Illusion [34]
2 years ago
8

{ question \hookleftarrow}" alt=" \sf \huge{ question \hookleftarrow}" align="absmiddle" class="latex-formula">
If \alpha \: and \: \beta are roots of a equation " ax² + by + c ", then find the value of the following in terms of a , b and c ~


\boxed{ \boxed{ \sf  \sqrt{ \alpha}  +   \sqrt{ \beta }  = \:  ?}}


​
Mathematics
2 answers:
BabaBlast [244]2 years ago
5 0

\underline{\bf{Given \:equation:-}}

\\ \sf{:}\dashrightarrow ax^2+by+c=0

\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.

\sf We\:know,

\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}

\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}

\underline{\large{\bf Identities\:used:-}}

\boxed{\sf (a+b)^2=a^2+2ab+b^2}

\boxed{\sf (√a)^2=a}

\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}

\boxed{\sf \sqrt{\sqrt{a}}=a}

\underline{\bf Final\: Solution:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}

\bull\sf Apply\: Squares

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}

\bull\sf Put\:values

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}

\bull\sf Simplify

\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}

\underline{\bf More\: simplification:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}

\underline{\Large{\bf Simplified\: Answer:-}}

\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}

GarryVolchara [31]2 years ago
5 0

The value  \sqrt{\alpha } +\sqrt{\beta } in terms of a, b and c is \sqrt{{(\frac{b}{a})^2 } +2\sqrt{\frac{c}{a} } } \\

<h3 />

Roots of a quadratic equation

Given the quadratic equation ax² + bx + c, the sum and product  of the roots are expressed as:

  • \alpha +\beta =-\frac{b}{a}
  • \alpha \beta =\frac{c}{a}

Get the value of the radical expression \sqrt{\alpha } +\sqrt{\beta }

Taking the square of the expression will give:

  • (\sqrt{\alpha } +\sqrt{\beta } )^2=(\sqrt{\alpha } )^2+(\sqrt{\beta } )^2+2\sqrt{\alpha \beta}

Take the square root of both sides:

\sqrt{(\sqrt{\alpha } +\sqrt{\beta } )^2}  =\sqrt{(\sqrt{\alpha } )^2+(\sqrt{\beta } )^2+2\sqrt{\alpha \beta} } \\&#10;\sqrt{\alpha } +\sqrt{\beta }=\sqrt{{(\alpha }+{\beta} )+2\sqrt{\alpha \beta} } \\

Substitute the product and the sum values into the expression to have:

\sqrt{\alpha } +\sqrt{\beta }=\sqrt{{(-\frac{b}{a})^2 } +2\sqrt{\frac{c}{a} } } \\\sqrt{\alpha } +\sqrt{\beta }=\sqrt{{(\frac{b}{a})^2 } +2\sqrt{\frac{c}{a} } } \\

Hence the value  \sqrt{\alpha } +\sqrt{\beta } in terms of a, b and c is \sqrt{{(\frac{b}{a})^2 } +2\sqrt{\frac{c}{a} } } \\

Learn more on the roots of equation here: brainly.com/question/25841119

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