Find the value of $k$ so that one root of the equations $3x^2+7x+(6-k)=0$ is $0$.

Question

Find the value of $k$ so that one root of the equations $3x^2+7x+(6-k)=0$ is $0$.

My Attempt: $$3x^2+7x+(6-k)=0$$

Now, $$\Delta = b^2-4ac$$ $$\Delta=(7)^2-4.(3).(6-k)$$ $$\Delta =49-72+12k$$ $$\Delta =12k-23$$

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Answer 1

Because $x=0$ is a root, it means when you take in $x=0$ the equation is true. Hence, just plug in $x=0$, and solve the equation $6-k=0$ to get $k=6$.

Answer 2

We can rewrite the equation as $$3x^2+7x+(6-k)=(x-0)(Ax+B) = x(Ax+B)$$ where $A$ and $B$ are numbers to be found, because $x=0$ is a root

We can then say $$3x^2+7x+(6-k)=Ax^2+Bx$$

We can equate coefficients to see that

\begin{align}3&=A\tag{$x^2$ term}\\ 7&=B\tag{$x$ term}\\ (6-k)&=0\tag{constant term}\\ &\Downarrow\\ 6&=k\end{align}

This method can be used for any value of the root $x$, not just $x=0$