Question

find a quadratic polynomial whose zeros are 2/3 & -7/3​

Answer 1

Answer:

$p(x)= 9x^{2} +15x-14$

Step-by-step explanation:

in the given question:

Let one zero be α= $\frac{2}{3}$

& other be β= $\frac{-7}{3}$

we know

Product of zeroes =α*β= $\frac{c}{a}$

                             = $\frac{2}{3}$*$\frac{-7}{3}$ = $\frac{-14}{9}$ (i)

Sum of zeroes = α+β = $\frac{-b}{a}$

                        = $\frac{2}{3}$ + $\frac{-7}{3}$ = $\frac{-5}{3}$ (ii)

now see

you will find that the value of a is different in eq(i)&(ii)

so to make them equal we will multiply eq(ii) by 3

⇒$\frac{-5}{3}$*$\frac{3}{3}$ = $\frac{-15}{9}$

Now

Standard form of any quadratic equation is

$p(x)=ax^{2} + bx+c;a\neq 0$

here

a = 9 ,- b= - 15(∵b=15)& c =  -14

∴The reqd eq is

$p(x)= 9x^{2} +15x-14$

Hope it is helpful