Review question

# If $7 - px - x^2 = 16 - (q + x)^2$, what are $p$ and $q$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9724

## Solution

Given that $y = 7 - px - x^2 = 16 - (q + x)^2$ for all values of $x$, where $p$ and $q$ are both positive,

1. calculate the values of $p$ and $q$,

By expanding the equation $7 - px - x^2 = 16 - (q + x)^2,$ we arrive at the equation $7 - px - x^2 = 16 - q^2 - 2qx - x^2.$ Since this is true for all $x$, we can equate coefficients, giving $7 = 16 - q^2$ and $-p = -2q$.

Thus $q^2=9$. Since the question tells us that $p$ and $q$ are both positive, we must have $q=3$, so $p=6$.

Thus $y=7-6x-x^2=16-(3+x)^2$.

1. state the maximum value of $y$ and the value of $x$ at which it occurs,

As $y = 16 - (3 + x)^2$, we see that the maximum value of $y$ is $16$, and this occurs when $x = -3$.

(This is because $(3 + x)^2 \ge 0$ no matter what the value of $x$, so $y$ is maximised when $(3 + x)^2=0$.)

1. find the range of values of $x$ for which $y$ is positive.

Again, the second representation is useful here: $y$ is positive when $16 - (3 + x)^2 > 0$, that is, when $(3 + x)^2 < 16$. This gives $-4 < 3 + x < 4$, so $-7<x<1$.

Alternatively, we can use the first representation, $y=7-6x-x^2=-(x^2+6x-7)=-(x-1)(x+7)$. This is zero when $x=1$ or $x=-7$. As this is a “vertex-up” parabola (since the coefficient of $x^2$ is negative), $y>0$ when $-7<x<1$.