If the ratio of roots of the equation $x^2+px+q=0$ is equal to the ratio of roots of the equation $x^2+bx+c=0$, then $p^2c-b^2q=$
- $0$
- $2$
- $\dfrac {-3}{2}$
- $-2$
Let the roots of x^2+px+q=0 be r1, r2 and x^2+bx+c=0 be r3, r4. Given r1/r2 = r3/r4. Using properties of roots, (r1+r2)^2 / (r1*r2) = p^2/q and (r3+r4)^2 / (r3*r4) = b^2/c. Since (r1+r2)^2 / (r1*r2) = (r1/r2 + r2/r1 + 2) and (r3/r4 + r4/r3 + 2) are equal, p^2/q = b^2/c, which implies p^2c = b^2q, so p^2c - b^2q = 0.
Let the roots of x squared plus px plus q equals 0 be alpha and beta, and let the roots of x squared plus bx plus c equals 0 be gamma and delta. Since the ratio of the roots is the same, alpha divided by beta equals gamma divided by delta, meaning the roots are proportional. By the properties of quadratic equations, if two equations have proportional roots, then the ratio of the sum of the roots to the product of the roots must be identical across both equations. This gives (negative p divided by q) equals (negative b divided by c), which simplifies to p times c equals b times q. Substituting this equality into the expression p squared c minus b squared q yields b squared q minus b squared q, which equals 0.