In a algebra quadratic question, it says:
The equation $3x^2+4x-k=0$ has two distinct real roots.
If 2 is a root of this equation, find the value of $k$ and the second root.
The first line of working for this question shows as:
$(3x+n)(x-2)=0$
The rest I more or less understand, but for this line of working, I get the $(x-2)$ because of a root at 2, but where does the $(3x+n)$ come from? What is $n$ and where does it come from?
Thank you
If you factor $$3x^2+4x-k$$ you know you'll have something of the form $$(ax+n)(bx+m)$$ assuming that there are two real roots, $x=-\frac{n}{a}$ and $x=-\frac{m}{b}$. The knowledge that $2$ is a root tells you that $b=1$ and $m=-2$. If we insert this, then we have $$(ax+n)(x-2)=3x^2+4x-k$$ $$ax^2+x(n-2a)x-2n=3x^2+4x-k\tag{1}$$ It should be clear from this that $a=3$ from the correspondence of the powers (there is only one term on each side that has $x^2$. We then have factored the original equation into $$(3x+n)(x-2)$$ Look at $(1)$ again, and note the correspondences: $$n-6=4$$ $$2n=k$$ Can you solve for $k$?
If $2$ is a root, it means $\;12+8-k=0$, hence $k=20$. The equation is thus $$3x^2+4x-20=0.$$ Using Vieta's relations, the product of the roots is $-\dfrac{20}3$, hence the other root is $-\dfrac{10}3$.
Alternatively, the sum of the roots is $-$, and the other ro43ot is $-\dfrac43-2=-\dfrac{10}3.$
We know that $x-2$ divides $3x^2+4x+k$ since $2$ is a root, so $3x^2+4x+k=(x-2)(ax+b)$ since we must have that $(ax)(x)=3x^2$, we know our $a$ must be 3. But without knowing the value of $k$ we cannot assume anything about $n$.
