Question

5. Given the quadratic equations 2x^2+px+9=0 and 3x^2-4x+q=0 have the same roots, find the value of p and of q. @ganeshie8

Answer 1

let the roots be r and s
First equation factors of ac that add up to p means factors of 18 and r + s = p
second equation factors of 3q that add up to -4 means r + s= -4

Answer 2

please note that:
for the equation:
\[2x ^{2}+px+9=0\]
I can write:
\[x _{1}+x _{2}=-\frac{ p }{ 2 },x _{1}*x _{2}=\frac{ 9 }{ 2 }\]
where x_1 and x_2 are the roots.
whereas for the second equation,
\[3y ^{2}-4y+q=0\]
we can write:
\[y _{1}+y _{2}=-\frac{ 4 }{ 3 },y _{1}*y _{2}=\frac{ q }{ 3 }\]
where y_1 and y_2 are the relative roots.
Now if we want that roots has to be equal then we can require that also the above systems have to be equal
so?

Answer 3

oops...\[y _{1}+y _{2}=\frac{ 4 }{ 3 }\]
sorry!

Answer 4

This one is actually much simpler - just divide the first equation by 2 and the second equation by 3 and then compare coefficients

Answer 5

@asnaseer because they have the same roots , thats why after you divide you can set the coefficients equal to each other.
equivalently, a quadratic polynomial with a coefficient of 1 for the x^2 term and which has roots r1,r2 is unique.