Question

the qudaratic polynomial whose zeros are 3/5 and -1/2 is
a)10x²-x-3
b) 10x²+x-3
c) 10x²-x+3
d) none of these​

Answer 1

a(alpha)=3/5

b(beta)= -1/2

a+b= 3/5 + (-1/2)=(6-5)/10= 1/10

ab=(3/5)(-1/2)= -3/10

WKT, p(x)= x^2 -(a+b)x + ab

= x^2 -(1/10)x -3/10

Now multiply 10 to RHS to remove denomenator

p(x) = 10x^2 -x -3

hence option (a)

hope it helps

Answer 2

Given:-

Two zeroes of the polynomial = $\sf{\dfrac{3}{5}}$ and $\sf{\dfrac{-1}{2}}$

To find:-

Quadratic polynomial.

Assumption:-

Let $\sf{\alpha = \dfrac{3}{5}}$ and $\sf{\beta = \dfrac{-1}{2}}$

Solution:-

$\sf{Sum\:of\:zeroes = \alpha + \beta}$

= $\sf{\alpha + \beta = \dfrac{3}{5} + \dfrac{-1}{2}}$

= $\sf{\alpha + \beta = \dfrac{2\times3 + (5\times -1)}{10}}$

= $\sf{\alpha + \beta = \dfrac{6-5}{10}}$

= $\sf{\alpha + \beta = \dfrac{1}{10}\longrightarrow[i]}$

Now,

$\sf{Product\:of\:zeroes = \alpha\times \beta}$

= $\sf{\alpha\beta = \dfrac{3}{5}\times \dfrac{-1}{2}}$

= $\sf{\alpha\beta = \dfrac{-3}{10}\longrightarrow[ii]}$

We know,

A Quadratic Equation is always in the form of:-

$\sf{x^2 + (\alpha + \beta)x + \alpha\beta = 0}$

Substituting the value of $\sf{\alpha + beta}$ and $\sf{\alpha\beta}$ from eq.[i] and [ii]

= $\sf{x^2 + \dfrac{1x}{10} + \dfrac{(-3)}{10}= 0}$

= $\sf{x^2 + \dfrac{x}{10} - \dfrac{3}{10}= 0}$

Taking LCM as 10

= $\sf{\dfrac{10x^2 + x - 3}{10} = 0}$

Multiplying 10 on both RHS and LHS

= $\sf{\dfrac{10x^2 + x - 3}{10}\times 10 = 0\times 10}$

= $\sf{10x^2 + x - 3 = 0}$

Therefore the quadratic equation is:-

${\boxed{\sf{10x^2 + x - 3}}}$

______________________________________

Verification:-

$\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

= $\sf{\dfrac{-1}{10} = \dfrac{-1}{10}}$ [Verified]

$\sf{Product\:of\:zeroes = \dfrac{Constant\:Term}{Coefficient\:of\:x^2}}$

= $\sf{\dfrac{-3}{10} = \dfrac{-3}{10}}$ [Verified]

______________________________________