Answer:
p = 2/3
q = 1/2
Step-by-step explanation:
The given equation is ,
[tex]\sf\to px^2 + px + 3q = 1 + 2x [/tex]
We can write it as ,
[tex]\sf\to px^2 + px + 3q - 1 -2x=0 [/tex]
Rearrange the terms ,
[tex]\sf\to px^2 - 2x + px + (3q -1)=0 [/tex]
This can be written as ,
[tex]\sf\to px^2 + x ( p - 2) + (3q -1) =0[/tex]
Now wrt Standard form of a quadratic equation ,
[tex]\bf \implies ax^2+bx + c = 0 [/tex]
we have ,
We know that product of zeroes :-
[tex]\to \sf q \times \dfrac{1}{p} = \dfrac{3q-1}{p } \\\\\sf\to 3q - 1 = q \\\\\sf\to 2q = 1 \\\\\sf\to \boxed{ q =\dfrac{1}{2}}[/tex]
Sum of roots :-
[tex]\to \sf q + \dfrac{1}{p} = \dfrac{2-p}{p} \\\\\sf\to \dfrac{ qp + 1}{p}= \dfrac{2-p}{p} \\\\\sf\to qp + 1 = 2 - p \\\\\sf\to p/2 + p = 1 \\\\\sf\to 3p/2 = 1 \\\\\sf\to \boxed{ p =\dfrac{2}{3}}[/tex]
