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secondary 3 | A Maths

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I need help for part b especially

Thank you!! I need it by 11pm

We need to start part ii “afresh”.

Suppose y = 0.

2x^2 + (h + 4)x + 2h = 0

Our discriminant of roots for this equation

= (h + 4)^2 - 4 (2) (2h)

= h^2 + 4h + 4h + 16 - 16h

= h^2 - 8h + 16

= (h - 4) (h - 4)

= (h - 4)^2

Wait for my next post.

As (h - 4)^2 is a squared expression, its lowest value is 0 (occurs when h = 4); for other values of h, (h - 4)^2 will spew out a positive value.

Note that squaring a negative number gets us a positive number, so it’s not possible for the square of a real number to be negative.

In other words,

- our discriminant = 0 when h = 4

- our discriminant > 0 for all other values of h

- “Two real and equal roots” when h = 4

- “Two real and distinct roots” for all other values of h

When an equation has two real and distinct roots, the corresponding curve will be two intercepts; when this happens, a graph will “cross over” from the positive side to the negative side (for smiley face curves) or the negative side to the positive side (for sad face curves).

Either way, the curve will exist on BOTH the positive and negative portions.

As we know it too well…

h = 4 is the only possible way.

Because only smiley face curves are capable of being “never negative”, we must ensure this.

Our coefficient is thankfully 2, so we have our answer, h = 4.

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