Math, asked by jibiskaria, 1 year ago

Please answer the 24th question. Correct answer will get marked brainliest.

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Answers

Answered by abhi569
4

Answer:

Required equation is 25x^2 - 30x + 4.

Required polynomial is 25x^2 - 30x + 4 .

Step-by-step explanation:

May be : Given method is not in your book.

According to this question :

Zeroes of an equation are ( 3 + √5 ) / 5 and ( 3 - √5 ) / 5 .

We know, every quadratic equation can be expressed in the form of ( x - a )( x - b ) = 0 , where a and b are the zeroes of that equation.

So, if ( 3 + √5 ) / 5 and ( 3 - √5 ) / 5 are zeroes :

= > { x - ( 3 + √5 ) / 5 }{ x - ( 3 - √5 ) / 5 } = 0

= > { ( 5x - 3 - √5 ) / 5 }{ ( 5x - 3 + √5 ) / 5 } = 0

= > ( 5x - 3 - √5 )( 5x - 3 + √5 ) = 0

= > 5x( 5x - 3 + √5 ) - 3( 5x - 3 + √5 ) - √5( 5x - 3 + √5 ) = 0

= > 25x^2 - 15x + 5√5 x - 15x + 9 - 3√5 - 5√5 x + 3√5 - 5 = 0

= > 25x^2 - 30x + 4 = 0

Hence the required equation is 25x^2 - 30x + 4 = 0.

Method 2

Given zeroes : ( 3 - √5 ) / 5 and ( 3 + √5 ) / 5 .

Sum of zeroes :

= > ( 3 - √5 ) / 5 + ( 3 + √5 ) / 5

= > ( 3 - √5 + 3 + √5 ) / 5

= > 6 / 5

Product of zeroes :

= > ( 3 - √5 ) / 5 \times ( 3 + √5 ) / 5

= > ( 3 - √5 )( 3 + √5 ) / 25

= > ( 9 - 5 ) / 25 { ( a + b )( a - b ) = a^2 - b^2 }

= > 4 / 25

From the properties of quadratic equations :

Quadratic equation : x^2 - ( sum of zeroes )x + product of zeroes = 0

= > x^2 - ( 6 / 5 )x + ( 4 / 25 ) = 0 ... { from above }

= > ( 25x^2 - 30x + 4 ) / 25 = 0

= > 25x^2 - 30x + 4 = 0

Hence the required equation is 25x^2 - 30x + 4 = 0.

And required polynomial is 25x^2 - 30x + 4 .


mysticd: It is a Polynomial not an equation , plz remove" =0"
mysticd: 25x²-30x+4 is correct
jibiskaria: Thank you so much for telling!
abhi569: @mysticd, I have added your one too. Satisfied ?
jibiskaria: Thank you so much @abhi569
jibiskaria: There's one slight mistake...you've written x^2+sx+p but it's actually x^2-sx+p
jibiskaria: I'm sure you did it by mistake.. but no problem. Thank you so much for answering!
abhi569: Welcome.
abhi569: :-)
jibiskaria: :)
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