The teacher has written an equation of the form $p(x) = 0$ on the board, where $p(x)$ is a quadratic, but Heather can't read the linear term. She can see that the quadratic term is $4x^2$ and that the constant is $-24$. She asks her neighbor, Noel, what the linear term is.
Noel decides to tease her and just says, ``One of the roots is 4.''
Heather then says, ``Oh, thanks!''
She then correctly writes down the linear term. What was the linear term? Write the entire term, not just the coefficient.
So
p(x) = 4x^2 + kx - 24
we also know p(4) = 0
4(4^2) + 4k - 24 = 0
solve for k and you have the linear term kx
Well, Heather seems to have a knack for solving quadratic equations with just one root. Maybe she has some magical powers!
Let's analyze the situation. We know that the quadratic term is $4x^2$ and the constant term is $-24$. Heather's helpful neighbor, Noel, decides to give her a clue and says that one of the roots is 4. So, we can conclude that the equation is of the form $p(x) = 4x^2 + ax - 24 = 0$, where $a$ represents the mysterious linear term we're looking for.
If one root of a quadratic equation is given, we know that the quadratic term must be split into two linear factors. Since one root is 4, the quadratic equation can be factored as $(x - 4)(4x + q)$, where $q$ represents the constant term of the unknown linear factor.
Now, let's multiply these two factors to get our quadratic equation back: $4x^2 + qx - 16x - 4q$.
Comparing this to our original quadratic equation $4x^2 + ax - 24$, we can see that $-16x$ must be equal to $ax$ and $-4q$ must be equal to $-24$. So, we have:
$ax - 16x = 0 \implies a - 16 = 0 \implies a = 16$
$-4q = -24 \implies q = 6$
Therefore, the missing linear term is $16x$. Heather can now accurately write down the entire term, thanks to Noel's clever little clue.
Let's assume the quadratic equation is of the form $p(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants.
Given that the quadratic term is $4x^2$ and the constant is $-24$, we have $p(x) = 4x^2 + bx - 24$.
Heather's neighbor, Noel, says that one of the roots is 4.
Knowing that one root is 4, we can use the fact that the sum of roots is equal to the negation of the coefficient of the linear term divided by the coefficient of the quadratic term. Mathematically, this can be written as:
Sum of roots = $-\frac{b}{a}$
Since one root is 4, we have:
Sum of roots = 4 + (Other Root) = $-\frac{b}{a}$
Substituting the known values into the equation above, we get:
4 + (Other Root) = $-\frac{b}{4}$
Simplifying further:
Other Root = $-\frac{b}{4} - 4$
Since a quadratic equation has two roots, the linear term of the equation can be found using the sum of roots. The sum of roots is equal to the negation of the coefficient of the linear term divided by the coefficient of the quadratic term.
Therefore, the linear term of the equation is $\boxed{\left(-\frac{b}{4} - 4\right)x}$.
To find the linear term, we need to determine the factored form of the quadratic equation.
First, let's recall the general form of a quadratic equation: $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants.
Given that the quadratic term is $4x^2$ and the constant term is $-24$, we can rewrite the equation as $4x^2 + bx - 24 = 0$.
Now, let's use the information provided by Noel that one of the roots is 4. This means that when $x = 4$, the equation evaluates to 0.
Substituting $x = 4$ into the equation, we get:
$4(4)^2 + b(4) - 24 = 0$
$64 + 4b - 24 = 0$
$4b + 40 = 0$
$4b = -40$
$b = -10$
Therefore, the linear term is $-10x$.