Question

If one of the roots of \[p{x^2} + qx + r = 0\] is thrice the other root then show that\[3{q^2} = 16pr\].

Answer 1

Hint: At first, we will find out the roots of the given quadratic equation. We will apply Sreedhar Acharya’s formula.
Let us consider, the general equation of quadratic equation is \[a{x^2} + bx + c = 0\]
The roots are, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Let us take, \[{r_1} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\] and \[{r_2} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\]
From the roots we get,
\[{r_1} + {r_2} = \dfrac{{ - b}}{{2a}}\], \[{r_1}.{r_2} = \dfrac{c}{a}\].
After finding the roots using the above method we will prove the given problem.

Complete step-by-step answer:
It is given that one of the roots of \[p{x^2} + qx + r = 0\] is thrice the other root.
Let us consider, \[\alpha ,\beta \] be the roots of the given equation.
Using the roots, the given equation can be written as,
\[{x^2} + (\alpha + \beta )x + \alpha \beta = 0\]
By using the relation between the roots and coefficients we get,
\[(\alpha + \beta ) = \dfrac{{ - q}}{p}\] and \[\alpha \beta = \dfrac{r}{p}\]
As per the statement given in the question, one root of the given equation is thrice of another root.
Therefore, we get, \[\alpha = 3\beta \]
Therefore by substituting \[\alpha = 3\beta \]in \[(\alpha + \beta ) = \dfrac{{ - q}}{p}\] we get,
\[4\beta = \dfrac{{ - q}}{p}\]
Now let us square the above equation, then we get,
\[16{\beta ^2} = \dfrac{{{q^2}}}{{{p^2}}}\]….. (1)
Therefore by substituting \[\alpha = 3\beta \]in \[\alpha \beta = \dfrac{r}{p}\] we get,
\[3{\beta ^2} = \dfrac{r}{p}\]
Simplifying the above equation, we get,
\[{\beta ^2} = \dfrac{r}{{3p}}\]…. (2)
From equation (1) and equation (2) we get,
\[\dfrac{{{q^2}}}{{16{p^2}}} = \dfrac{r}{{3p}}\]
Simplifying the above equation by cross multiplying the terms in the above equation, we get,
\[3{q^2} = 16pr\]
Hence, we have proved the required result.

Note:
The nature of the roots depends on the discriminant part that is \[\sqrt {{b^2} - 4ac} \].
If the discriminant is less than zero, the roots will be imaginary.
If the discriminant is greater than and equals zero, the roots will be real.