Question

find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and their co efficients 2s^2+5s+3​

Answer 1

Given Quadratic polynomial :- 2s²+5s+3

To find :- Zeroes of the given Quadrilateral and verify the relationship between the zeroes and coefficients of the polynomial .

Used Concepts :- For a given Quadratic Polynomial ax²+bx+c .

The sum of roots are given by = -b/a

The Product of roots are given by = c/a .

Solution :-

p(s) = 2s²+5s+3

For Zeroes , p(s) = 0

Therefore, 2s²+5s+3 = 0

2s²+2s+3s+3 = 0

2s ( s + 1 ) + 3 ( s + 1 ) = 0

( 2s + 3 ) ( s + 1 ) = 0

Either , 2s + 3 = 0 or s + 1 = 0

2s = -3 s = -1

s = -3/2 s = -1

Therefore , s = -3/2 , -1

Now , Sum of Zeroes :-

$\frac{ - 3}{2} + ( - 1)$

$\frac{ - 3}{2} - 1$

$\frac{ - 3 - 2}{2}$

$\frac{ - 5}{2}$

Now Sum of Zeroes by -b/a .

2s² + 5s + 3 = 0

Here , a = 2 , b = 5 , c = 3

Now , Sum of Zeroes :-

-b/a

-5/ 2 .

Now , Product of roots ,

-3/2 × -1

3/2

Now , Product of roots by c/a , we get ,

c/a

3/2 .

Hence , Verified !!

Answer 2

let f(x) = $2s^2+5s+3$

=> $2s^2+5s+3$

=> $2s^2 +s(2+3)+3$

=> $2s^2 +2s+3s+3$

=> $2s(s +1)+3(s+1)$

=> $(s +1)(2s+3)$

then,

f (x) = 0

so, if (s + 1) = 0

   s = -1

if,  2s + 3 = 0

      s = $\frac{-3}{2}$

zeroes of polynomial  = $-1,\frac{-3}{2}$

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Now,

the relationship between the zeroes and their coefficients

=> $sum\ of \ zeroes=\frac {-(coefficient \ of s)}{coefficient \ of s^2}$

=> $-1+(\frac{-3}{2}) =\frac {-(5)}{2}$

=> $\frac{-2-3}{2} =\frac {-5}{2}$

=> $\frac{-5}{2} =\frac {-5}{2}$

and,

=> $product \ of \ zeros= \frac{constant \ term}{cofficient of s^2}$

=> $(-1)(\frac{-3}{2} )= \frac{3}{2}$

=> $\frac{3}{2} = \frac{3}{2}$