Math, asked by rohansaivijjapu, 4 months ago

the representation of hanging wires on the bridge whose sum of the zeroes is -3 and the product of zeroes is 5 is ​

Answers

Answered by Swarup1998
6

POLYNOMIALS AND ZEROES

If l be the sum of zeroes and m be the product of zeroes of the polynomial f(x), then it is given by

\quad f(x)=x^{2}-lx+m

Step-by-step explanation:

Given that the sum of zeroes is (-3) and the product of zeroes is 5.

Comparing with f(x), we get

  • l=-3 and m=5.

Putting the values in f(x), we obtain the required polynomial

\quad f(x)=x^{2}+3x+5

Answer:

Therefore the representation of hanging wires on the bridge is given by x^{2}+3x+5.

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Answered by pulakmath007
10

SOLUTION

TO DETERMINE

The representation of hanging wires on the bridge whose sum of the zeroes is -3 and the product of zeroes is 5

CONCEPT TO BE IMPLEMENTED

If sum of the zeroes and product of the zeroes of a polynomial f(x) are given then the quadratic polynomial is given by

 \sf{f(x) =  {x}^{2}  - (Sum  \: of \:  the zeroes)x +  Product  \: of  \: the \:  Zeroes }

EVALUATION

Here it is given that sum of the zeroes is -3 and the product of zeroes is 5

So the required polynomial is

 \sf{f(x) =  {x}^{2}  - (Sum  \: of \:  the zeroes)x +  Product  \: of  \: the \:  Zeroes }

 \implies \sf{f(x) =  {x}^{2}  -( - 3)x + 5 }

 \implies \sf{f(x) =  {x}^{2}   + 3x + 5 }

FINAL ANSWER

The representation of hanging wires on the bridge whose sum of the zeroes is -3 and the product of zeroes is 5 is

 \sf{f(x) =  {x}^{2}   + 3x + 5 }

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Additional Information :-

A general equation of quadratic equation is

a {x}^{2} +  bx + c = 0

Now one of the way to solve this equation is by SRIDHAR ACHARYYA formula

For any quadratic equation

a {x}^{2} +  bx + c = 0

The roots are given by

 \displaystyle \: x =  \frac{ - b \pm \:  \sqrt{ {b}^{2} - 4ac } }{2a}

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