Math, asked by sushantodey765, 2 months ago

find the roots of the following equations​

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Answers

Answered by ItsPrithvirajChauhan
3

Answer:

L.H.S

= √2x² + 7x + 5√2.

= √2x² + 2x + 5x + 5√2.

= (√2x² + 2x) + ( 5x + 5√2).

=. ( √2x² + √2 × √2 × x) + ( 5x + 5√2).

=. √2x( x + √2 ) + 5( x + √2 ).

= ( x + √2 ) ( √2x + 5 )

1. x + √2 = 0

x = - √2

2. √2x + 5 = 0

√2x = - 5

x = -5/√2.

Step-by-step explanation:

solve by splitting the middle term.

Answered by ImperialGladiator
7

Answer:

  • \rm -\sqrt{2} \:  and \:\dfrac{ - 5\sqrt{2}}{ {2} }

Explanation:

Given equation,

 =  \boldsymbol { \sqrt{2} {x}^{2}  + 7x + 5 \sqrt{2} = 0  }

On comparing with the general form of a quadratic equation \boldsymbol {ax^2 + bx + c = 0}

We get,

  • \boldsymbol {a = \sqrt{2}}
  • \boldsymbol {b = 7}
  • \boldsymbol {c = 5\sqrt{2}}

Using quadratic formula,

 \implies \:  \boldsymbol{x =  \dfrac{ - b \pm  \sqrt{ {b}^{2}  - 4ac} }{2a} }

\implies \:  x =  \dfrac{ - 7 \pm  \sqrt{ {(7)}^{2}  - 4( \sqrt{2})(5 \sqrt{2}) } }{2 \sqrt{2} }

\implies \:  x =  \dfrac{ - 7 \pm  \sqrt{ 49  - 40 }}{2 \sqrt{2} }

\implies \:  x =  \dfrac{ - 7 \pm  \sqrt{9}}{2 \sqrt{2} }

\implies \:  x =  \dfrac{ - 7 \pm 3}{2 \sqrt{2} }

\implies \:  x =  \dfrac{ - 7  +  3}{2 \sqrt{2} }  { \rm \: or} \:  \dfrac{ - 7 - 3}{2 \sqrt{2} }

\implies \:  x =  \dfrac{ - 4}{2 \sqrt{2} }  \:  { \rm \: or} \:  \dfrac{ - 10}{2 \sqrt{2} }

\implies \:  x =  \dfrac{-2}{\sqrt{2}}  \:  { \rm \: or} \:  \dfrac{ - 5}{ \sqrt{2} }

On rationalising the denominator,

\implies \:  x =  -\sqrt{2} \:  { \rm \: or} \:  \dfrac{ - 5\sqrt{2}}{ {2} }

{ \underline {\sf {\therefore{Roots \: of \: the \: equation \: are \: -\sqrt{2} \:  and \:\dfrac{ - 5\sqrt{2}}{ {2} } }}}}

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