- #1

- 10

- 0

5+4x -x^2 = m - (x - n)^2

for all real values of x.

If someone could work this out, Then ill show my workings and ask questions (its to do with the arrangements of N^2 after expanding brackets mainly.

Thanks in advanced,

ASMATHSHELPME

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- Thread starter ASMATHSHELPME
- Start date

- #1

- 10

- 0

5+4x -x^2 = m - (x - n)^2

for all real values of x.

If someone could work this out, Then ill show my workings and ask questions (its to do with the arrangements of N^2 after expanding brackets mainly.

Thanks in advanced,

ASMATHSHELPME

- #2

AKG

Science Advisor

Homework Helper

- 2,565

- 4

ax² + bx + c

= a[x² + (b/a)x] + c

= a[x² + (b/a)x + (b/2a)² - (b/2a)²] + c

= a[x² + (b/a)x + (b/2a)²] - a(b/2a)² + c

= a[x + (b/2a)]² + (c - b²/4a)

Complete the square for the left side of your equation, and it will look like the right side, but instead of m and n, you'll have numbers, and those are obviously the numbers you're trying to find.

- #3

- 319

- 0

multiply both sides by -1 , to get, x^2-4x-5 which can be written as (x-2)^2-9 , which is in the form (x - n)^2 + m

so n=2, m=-9

- #4

VietDao29

Homework Helper

- 1,424

- 3

Recheck that, roger. The problem states: 5+4x -x^2 = m - (x - n)^2, not 5+4x -x^2 = m + (x - n)^2roger said:so n=2, m=-9

Viet Dao,

- #5

- 319

- 0

whoops,

m=9

m=9

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