Question

Jessie has rewritten the quadratic equation x2-32x +54 -0 in the form (-p) = 202. What is the value of p in the rewritten equation?

Answer 1

Given:

The quadratic equation is:

$x^2-32x+54=0$

It can be written as $-p=202$.

To find:

The value of p in the rewritten equation.

Solution:

We have,

$x^2-32x+54=0$

Isolate the constant term.

We need to make 202 on the right side. So, add 256 on both sides.

$x^2-32x+256=-54+256$

$x^2-32x+256=202$

$-(-x^2+32x-256)=202$

Let $-x^2+32x-256=p$, then

$-p=202$

Therefore, the value of p is $-x^2+32x-256$.

The given equation can be written as:

$54=-x^2+32x$

Adding 148 on both sides, we get

$54+148=-x^2+32x+148$

$202=-(x^2-32x-148)$

Let $x^2-32x-148=p$, then

$202=-p$

Therefore, the another possible value of p is $x^2-32x-148$.