Math, asked by pareekdp2005, 10 months ago

Find the zeros of the polynomial x^2+1/6x-2 and verify the relationship between its zeros and coefficients of the polynomial​

Answers

Answered by Anonymous
32

Solution :

\bf{\underline{\bf{Given\::}}}}

The polynomial x² + 1/6x - 2.

\bf{\underline{\bf{To\:find\::}}}}}

The zeroes and verify the relationship between it's zeroes and coefficient of the polynomial.

\bf{\underline{\bf{Explanation\::}}}}}

We have p(x) = x² + 1/6x - 2

Zero of the polynomial is p(x) = 0

So;

\mapsto\sf{x^{2} +\dfrac{1}{6} x-2=0}\\\\\mapsto\sf{6x^{2} +x-12=0}\\\\\mapsto\sf{6x^{2} +9x-8x-12=0}\\\\\mapsto\sf{3x(2x+3)-4(2x+3)=0}\\\\\mapsto\sf{(2x+3)(3x-4)=0}\\\\\mapsto\sf{2x+3=0\:\:Or\:\:3x-4=0}\\\\\mapsto\sf{2x=-3\:\:Or\:\:3x=4}\\\\\mapsto\sf{\green{x=-\dfrac{3}{2} \:\:Or\:\:x=\dfrac{4}{3} }}

∴ The α = -3/2 and β = 4/3 are the zeroes of the polynomial.

As the given quadratic polynomial as we compared with ax² + bx + c

  • a = 6
  • b = 1
  • c = -2

So;

\red{\underline{\underline{\bf{Sum\:of\:the\:zeroes\::}}}}

\longrightarrow\sf{\alpha +\beta=\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x) }{Coefficient\:of\:(x)^{2}} }\\\\\\\longrightarrow\sf{\dfrac{-3}{2} +\dfrac{4}{3} =\dfrac{-1}{6} }\\\\\\\longrightarrow\sf{\dfrac{-9+8}{6} =-\dfrac{1}{6}} \\\\\\\longrightarrow\sf{\green{\dfrac{-1}{6} =\dfrac{-1}{6} }}

\red{\underline{\underline{\bf{Product\:of\:the\:zeroes\::}}}}

\longrightarrow\sf{\alpha \times \beta=\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:(x)^{2}} }\\\\\\\longrightarrow\sf{\dfrac{-3}{2} \times\dfrac{4}{3} =\dfrac{-2}{1} }\\\\\\\longrightarrow\sf{\cancel{\dfrac{-12}{6}} =-2} \\\\\\\longrightarrow\sf{\green{-2 =-2 }}

Thus;

Relationship between zeroes and coefficient is verified .

Answered by kiran01486
18

Answer:

Step-by-step explanation:

Here the given equation is

x²+(1/6x)-2=0

multiplying the above equation by 6 we get

6x²+x-12=0

or,6x²+9x-8x-12=0

or,3x(2x+3)-4(2x+3)=0

or,(3x-4)(2x+3)=0

Hence the zeros of the polynomial are

x=4/3 andx=-3/2

Now for a quadratic polynomial

ax²+bx+c=0

the relation between coefficient a b and c and their roots m and n are

m+n=-b/a

mn=c/a

For the given equation

m=4/3 n=-3/2

and a=1,b=1/6 and c=-2

now m+n=-1/6

and -b/a=-1/6

therefore,m+n=-b/a

also m x n=-2

and c/a=-2

Hence m x n=c/a

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