Question

find relationship between zeros and coefficient 7y2_11/3y_2/3​

Answer 1

Answer:

Given:

• polynomial 7y² - 11/3y - 2/3
• To find the relation of zeros and polynomial.

Solving Question:

  we are given the polynomial,so first we have to find the zeros then find out the relation between zeros and coefficients.

Solution:

 $7y^2-\dfrac{11}{3}y-\dfrac{2}{3}=0\\\\Multiply\:with\:\:3\\\\\implies 21y^2-11y-2=0(we\:will\:consider\:this)\\\\to\:find\:zeros\\\\quadratic\:formula =  \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\\\substitue\:the\:values\\\\ a= 21;b=-11;c=2\\\\\dfrac{-(-11)\pm \sqrt{121-4(-2)(21)}}{2*21}\\\\or, \dfrac{11 \pm \sqrt{289}}{42}\\\\\dfrac{11\pm 17}{42} \implies zeeros \:are \dfrac{28}{42} \implies \dfrac{2}{3} \:and \\\dfrac{11-17}{42} \implies \dfrac{6}{42} \implies \dfrac{-1}{7}$

Zeros are 2/3 and -1/7

Then, to find relations

we know that ,

α +β = -b/a

α * β = c/a

where 'α' and 'β' are the roots.

to verify it

$sum\:of\:roots\\\\=\dfrac{2}{3}+\dfrac{-1}{7} \implies \dfrac{14-3}{21} \:[taking\:L.C.M]\:\implies \dfrac{11}{21}$ .........(L.H.S)

to find, -b/a  

-b/a = -(-11)/21 = 11/21 .....(R.H.S)

hence , R.H.S = L.H.S

then, the product

$poduct= \dfrac{2}{3}*\dfrac{-1}{7}=\dfrac{-2}{21}$........(L.H.S)

α * β = c/a

to find ,c/a

c/a = -2/21    

Hence this is the relationship

α +β = -b/a

α * β = c/a

where 'α' and 'β' are the roots.