Quadratic Roots Sum Practice

• From the first quadratic equation: $5x^2 + 15x + a$ , the sum of roots will be: $\boxed{\color{#D61F06}{p+q = \frac{-15}{5} \implies (-3)}}$

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

$\begin{aligned} \dfrac{-b}{1} &= -\bigg(p +q \bigg) - \bigg(\dfrac{p}{3} + \dfrac{q}{3}\bigg) \\ &= - \bigg(\dfrac{4p}{3} + \dfrac{4q}{3}\bigg) \\ &= \dfrac{-4}{3} \big({\color{#D61F06}{p + q}}\big) \\ &= \dfrac{-4}{3} \big({\color{#D61F06}{-3}}\big) \\ \\ -b &= +4 \\ b &= \boxed{-4} \end{aligned}$

We start by examining the first equation: 5x^2+15x+a

The only piece of information we get from this equation is p+q=-3 (By the sum of the roots of a quadratic)

Then we look at the other quadratic: x^2+bx+c

We can see that the value of b is the opposite of the sum of the roots of the equation.

We are given that the roots of the equation are -(p+q) and -(p/3+q/3)

Adding these together we have: -(p+q+p+q/3)=-b

Plugging in p+q=-3 we have: -[-3+(-1)]=-b

Solving the b we get b=-4.

Applying Vieta's formula on the first quadratic equation $p+q=\frac{-15}{5}=-3$ Applying Vieta's formula on the second quadratic equation $-(p+q)+[-(\frac{p}{3}+\frac{q}{3})]=-p-q-\frac{p}{3}-\frac{q}{3}=\frac{-3p-3q-p-q}{3}=\frac{-4p-4q}{3}=\frac{\cancel{-}4(p+q)}{3}=\cancel{-}b$ Putting value of $p+q$ $\frac{4(-\cancel{3})}{\cancel{3}}=b$ $\boxed{b=-4}$