Quadratic Roots Product Practice

• From the first quadratic equation: $3x^2 + ax + 6$ , the product of roots will be: $\boxed{\color{#D61F06}{pq = \frac{6}{3} \implies 2}}$

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• From the second quadratic equation: $x^2 + bx + c$ , the product of roots will be:

\[\begin{align} \dfrac{c}{1} &= \bigg(p + \dfrac{1}{q}\bigg) \cdot \bigg(q + \dfrac{1}{p}\bigg) \\ &= {\color{Red}{pq}} + 1 + 1 + \dfrac{1}{\color{Red}{pq}} \\ &= {\color{Red}{2}} + 2 + \dfrac{1}{{\color{Red}{2}} } \\ &= \boxed{4 \dfrac{1}{2}}

\end{align}\]

We first look at the first equation 3x^2+ax+6.

The product of the roots of the equation 3x^2+ax+6 is 2.

Now we find the product of the roots of the other equation.

We expand (p+1/q)(q+1/p) to get pq+2+1/pq

So the product of the roots of the equation is 2+2+1/2=4 1/2

So the value of c is 4 1/2.

Applying Vieta's formula on the first equation $\boxed{pq=\frac{6}{3}=2}$ Applying Vieta's formula on the second equation $(p+\frac{1}{q})(q+\frac{1}{p})=pq+1+1\frac{1}{pq}=pq+\frac{1}{pq}+2=c$ Putting value of $pq$ $2+\frac{1}{2}+2=\boxed{4\frac{1}{2}=c}$