If one root of the equation $ax^2 + bx + c = 0;> a,b,c \in R$ is cube of other, then $\displaystyle \frac{b^2-2ac}{a+c}$ is equal to
- $\pm \sqrt{ac}$
- $\pm \sqrt{bc}$
- $\pm \dfrac{c^2}{a}$
- $\pm \dfrac{c^2}{b}$
Let roots be alpha and alpha^3. Sum = -b/a = alpha + alpha^3. Product = c/a = alpha^4. Then alpha = (c/a)^(1/4). Substitute into sum equation to relate a, b, c. The expression (b^2 - 2ac)/(a+c) simplifies to +/- sqrt(ac).
Let the roots be alpha and alpha^3, so their product gives c divided by a = alpha^4, meaning alpha = plus or minus the fourth root of (c divided by a). The sum of the roots is alpha + alpha^3 = -(b divided by a), so we can write alpha(1 + alpha^2) = -(b divided by a). Squaring both sides gives alpha^2(1 + alpha^2)^2 = (b^2 divided by a^2). Expanding the left side yields alpha^2 + 2alpha^4 + alpha^6 = (b^2 divided by a^2). Substituting alpha^2 = the square root of (ac) divided by a and alpha^4 = c divided by a simplifies the expression to (2c divided by a) + 2(square root of ac) divided by a, which equals (b^2 - 2ac) divided by a^2. Thus, (b^2 - 2ac) divided by (a + c) simplifies to plus or minus the square root of ac.