Fast Compute Grants

[ASSESS]

We are given two quadratic polynomials $P(x)$ and $Q(x)$ with leading coefficients $2$ and $-2$, respectively. The graphs of both polynomials pass through the points $(16,54)$ and $(20,53)$. We need to find the value of $P(0) + Q(0)$.

[ADVANCE]

Let's express the quadratic polynomials $P(x)$ and $Q(x)$ in the general form:
$ P(x) = 2x^2 + bx + c $
$ Q(x) = -2x^2 + dx + e $

Since the graphs of both polynomials pass through the points $(16,54)$ and $(20,53)$, we can substitute these points into the equations of $P(x)$ and $Q(x)$ to form a system of equations.

For $P(x)$:
$ P(16) = 2(16)^2 + 16b + c = 54 $
$ 512 + 16b + c = 54 $
$ 16b + c = -458 \quad \text{(Equation 1)} $

$ P(20) = 2(20)^2 + 20b + c = 53 $
$ 800 + 20b + c = 53 $
$ 20b + c = -747 \quad \text{(Equation 2)} $

For $Q(x)$:
$ Q(16) = -2(16)^2 + 16d + e = 54 $
$ -512 + 16d + e = 54 $
$ 16d + e = 566 \quad \text{(Equation 3)} $

$ Q(20) = -2(20)^2 + 20d + e = 53 $
$ -800 + 20d + e = 53 $
$ 20d + e = 853 \quad \text{(Equation 4)} $

[ADVANCE]

Now, we can solve the system of equations for $P(x)$:
Subtract Equation 1 from Equation 2:
$ (20b + c) - (16b + c) = -747 - (-458) $
$ 4b = -289 $
$ b = -\frac{289}{4} $

Substitute $b = -\frac{289}{4}$ into Equation 1:
$ 16\left(-\frac{289}{4}\right) + c = -458 $
$ -1156 + c = -458 $
$ c = 698 $

So, for $P(x)$, we have:
$ P(x) = 2x^2 - \frac{289}{4}x + 698 $
$ P(0) = 698 $

Next, we can solve the system of equations for $Q(x)$:
Subtract Equation 3 from Equation 4:
$ (20d + e) - (16d + e) = 853 - 566 $
$ 4d = 287 $
$ d = \frac{287}{4} $

Substitute $d = \frac{287}{4}$ into Equation 3:
$ 16\left(\frac{287}{4}\right) + e = 566 $
$ 1148 + e = 566 $
$ e = -582 $

So, for $Q(x)$, we have:
$ Q(x) = -2x^2 + \frac{287}{4}x - 582 $
$ Q(0) = -582 $

[ADVANCE]

Finally, we can find $P(0) + Q(0)$:
$ P(0) + Q(0) = 698 + (-582) = 116 $

So, the value of $P(0) + Q(0)$ is $\boxed{116}$.

So I've got this problem here about quadratic polynomials $P(x)$ and $Q(x)$. $P(x)$ has a leading coefficient of $2$, and $Q(x)$ has a leading coefficient of $-2$. Both polynomials pass through the points $(16,54)$ and $(20,53)$. I need to find $P(0) + Q(0)$.

First, since both $P(x)$ and $Q(x)$ are quadratic polynomials, I can express them in the general form:

$P(x) = 2x^2 + bx + c$

$Q(x) = -2x^2 + dx + e$

Wait, actually, the leading coefficient is the coefficient of the highest power of $x$, which for a quadratic is $x^2$. So yes, $P(x) = 2x^2 + bx + c$ and $Q(x) = -2x^2 + dx + e$.

Now, both polynomials pass through the points $(16,54)$ and $(20,53)$. That means when $x=16$, $P(16)=54$, and when $x=20$, $P(20)=53$. Similarly for $Q(x)$.

So, I can set up equations using these points.

First, for $P(x)$:

$P(16) = 2(16)^2 + b16 + c = 54$

Similarly, $P(20) = 2(20)^2 + b20 + c = 53$

Let me calculate $16^2$ and $20^2$:

$16^2 = 256$

$20^2 = 400$

So,

$P(16): 2256 + 16b + c = 54 \Rightarrow 512 + 16b + c = 54 \Rightarrow 16b + c = 54 - 512 \Rightarrow 16b + c = -458$

$P(20): 2400 + 20b + c = 53 \Rightarrow 800 + 20b + c = 53 \Rightarrow 20b + c = 53 - 800 \Rightarrow 20b + c = -747$

Now, I have two equations:

1) $16b + c = -458$

2) $20b + c = -747$

I can subtract equation 1 from equation 2 to eliminate $c$:

$(20b + c) - (16b + c) = -747 - (-458)$

$20b + c - 16b - c = -747 + 458$

$4b = -289$

So, $b = -289 / 4 = -72.25$

Wait, but coefficients are usually integers, but maybe not. Let me check.

Now, plug $b$ back into one of the equations to find $c$. Let's use equation 1:

$16(-72.25) + c = -458$

$-1156 + c = -458$

$c = -458 + 1156 = 698$

So, $P(x) = 2x^2 - 72.25x + 698$

Hmm, $P(0)$ would be $c$, which is $698$.

Now, let's do the same for $Q(x)$.

$Q(16) = -2*(16)^2 + d16 + e = 54 \Rightarrow -2256 + 16d + e = 54 \Rightarrow -512 + 16d + e = 54 \Rightarrow 16d + e = 54 + 512 \Rightarrow 16d + e = 566$

$Q(20) = -2*(20)^2 + d20 + e = 53 \Rightarrow -2400 + 20d + e = 53 \Rightarrow -800 + 20d + e = 53 \Rightarrow 20d + e = 53 + 800 \Rightarrow 20d + e = 853$

Now, equations for $Q(x)$:

1) $16d + e = 566$

2) $20d + e = 853$

Subtract equation 1 from equation 2:

$(20d + e) - (16d + e) = 853 - 566$

$4d = 287$

$d = 287 / 4 = 71.75$

Now, plug $d$ back into equation 1:

$1671.75 + e = 566 \Rightarrow 1148 + e = 566 \Rightarrow e = 566 - 1148 = -582$

So, $Q(x) = -2x^2 + 71.75x - 582$

Therefore, $Q(0) = e = -582$

Now, $P(0) + Q(0) = 698 + (-582) = 116$

Wait, but the problem says "present the answer in LaTeX format: \boxed{Your answer}"

So, I should write $\boxed{116}$

But let me double-check my calculations to make sure I didn't make any mistakes.

First, for $P(x)$:

$P(16) = 2(256) + 16b + c = 512 + 16b + c = 54 \Rightarrow 16b + c = -458$

$P(20) = 2*(400) + 20b + c = 800 + 20b + c = 53 \Rightarrow 20b + c = -747$

Subtracting: $4b = -289 \Rightarrow b = -72.25$

Then, $c = -458 - 16*(-72.25) = -458 + 1156 = 698$

Similarly for $Q(x)$:

$Q(16) = -2*(256) + 16d + e = -512 + 16d + e = 54 \Rightarrow 16d + e = 566$

$Q(20) = -2*(400) + 20d + e = -800 + 20d + e = 53 \Rightarrow 20d + e = 853$

Subtracting: $4d = 287 \Rightarrow d = 71.75$

Then, $e = 566 - 16*71.75 = 566 - 1148 = -582$

So, $P(0) = 698$, $Q(0) = -582$, sum is $116$.

Seems correct.

Final Answer

$\boxed{116}$

So I have two quadratic polynomials, $P(x)$ and $Q(x)$, with leading coefficients 2 and -2, respectively. Both pass through the points $(16,54)$ and $(20,53)$. I need to find $P(0) + Q(0)$.

First, since both are quadratic polynomials, I can express them in the general form:

$P(x) = 2x² + bx + c$

$Q(x) = -2x² + dx + e$

But since both pass through the points $(16,54)$ and $(20,53)$, I can set up equations using these points.

For $P(x)$:

$P(16) = 2*(16)² + b \cdot 16 + c = 54$

$P(20) = 2(20)² + b \cdot 20 + c = 53$

Similarly for $Q(x)$:

$Q(16) = -2(16)² + d \cdot 16 + e = 54$

$Q(20) = -2(20)² + d \cdot 20 + e = 53$

Let's compute the squares:

$16² = 256$

$20² = 400$

So for $P(x)$:

$2256 + 16b + c = 54 \Rightarrow 512 + 16b + c = 54 \Rightarrow 16b + c = 54 - 512 = -458 \quad ...(1)$

$2400 + 20b + c = 53 \Rightarrow 800 + 20b + c = 53 \Rightarrow 20b + c = 53 - 800 = -747 \quad ...(2)$

For $Q(x)$:

$-2256 + 16d + e = 54 \Rightarrow -512 + 16d + e = 54 \Rightarrow 16d + e = 54 + 512 = 566 \quad ...(3)$

$-2400 + 20d + e = 53 \Rightarrow -800 + 20d + e = 53 \Rightarrow 20d + e = 53 + 800 = 853 \quad ...(4)$

Now, I have two systems of linear equations:

For $P(x)$:

$16b + c = -458 \quad ...(1)$

$20b + c = -747 \quad ...(2)$

Subtract equation (1) from equation (2):

$(20b + c) - (16b + c) = -747 - (-458)$

$4b = -289 \Rightarrow b = -\frac{289}{4}$

Now plug $b$ back into equation (1):

$16\left(-\frac{289}{4}\right) + c = -458 \Rightarrow -1156 + c = -458 \Rightarrow c = -458 + 1156 = 698$

So $P(x) = 2x² - \frac{289}{4}x + 698$

For $Q(x)$:

$16d + e = 566 \quad ...(3)$

$20d + e = 853 \quad ...(4)$

Subtract equation (3) from equation (4):

$(20d + e) - (16d + e) = 853 - 566$

$4d = 287 \Rightarrow d = \frac{287}{4}$

Now plug $d$ back into equation (3):

$16\left(\frac{287}{4}\right) + e = 566 \Rightarrow 1148 + e = 566 \Rightarrow e = 566 -1148= -582$

So $Q(x)=-2x^{2}+\frac{287}{4}x-582$.

Now, $P(0)=c=698$.

$Q(0)=e=-582$.

Therefore, $P(0)+Q(0)=698+(-582)=116$.

Final Answer

$\boxed{116}$.