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`.