How - 22 turn into 121....and in line seven how i get 226?

[Murshed]

Hi

I'm new just here so

In four number line, how minus 22 turn into 121 and number line how get 226 please?

 

[topsquark]

Hi

I'm new just here so

In four number line, how minus 22 turn into 121 and number line how get 226 please?View attachment 36610

The 22 didn't turn into 121. [imath]\Delta[/imath] is the discriminant (actually, 1/4 of the discriminant)
[imath]a r^2 + b r + c = 0 \implies \Delta = b^2 - 4 a c[/imath]

So, find [imath]\Detla[/imath] and divide by 4 and you get line 3.

The 226 is found by expanding the [imath]3^2 + (11 - 4 \sqrt{6})^2 = 3^2 + 11^2 - 88 \sqrt{6} + 96[/imath].

Just a thought... I'm getting the idea that you haven't learned how to solve Quadratic equations, nor how to FOIL out an expression?

-Dan

 

[stapel]

Hi

I'm new just here so

In four number line, how minus 22 turn into 121 and number line how get 226 please?View attachment 36610

What is the meaning of [imath]\Delta'[/imath]? Does it perhaps stand for "the value of the discriminant"? If so, then your value of "121" cannot be correct. (It is 22 that is being squared, not 11.) The value of the square root of [imath]\Delta'[/imath] is correct, however.

What were the instructions for this drawing? Are you supposed to be finding the *shaded* area?

Thank you!

 

[Dr.Peterson]

Hi

I'm new just here so

In four number line, how minus 22 turn into 121 and number line how get 226 please?View attachment 36610

As I read this, they are using [imath]\Delta'[/imath] to mean the discriminant divided by 4, namely [math]\Delta'=\frac{\Delta}{4}=\frac{b^2-4ac}{4}=\frac{b^2}{4}-ac=\left(\frac{b}{2}\right)^2-ac[/math]
Therefore [math]\Delta'=\left(\frac{22}{2}\right)^2-1\cdot25=11^2-25=121-25=96[/math]
That explains line 4.

As for the last line, evidently [imath]\mathscr{A}[/imath] means the shaded area.

The part in brackets expands to [math]3^2+\left(11-4\sqrt{6}\right)^2=3^2+11^2+\left(4\sqrt{6}\right)^2-2\cdot11\cdot 4\sqrt{6}=9+121+96-88\sqrt{6}=226-88\sqrt{6}[/math]

 

[Steven G]

b²-4ac = 22² - 4*1*25 = (2*11)² - 4*25 = 4*121-4*25 = 4(121-25) = 4(96). That is why they use Δ′ as Δ/4

This is a perfect case that shows that one needs to define their variables.

 

[Murshed]

The 22 didn't turn into 121. [imath]\Delta[/imath] is the discriminant (actually, 1/4 of the discriminant)
[imath]a r^2 + b r + c = 0 \implies \Delta = b^2 - 4 a c[/imath]

So, find [imath]\Detla[/imath] and divide by 4 and you get line 3.

The 226 is found by expanding the [imath]3^2 + (11 - 4 \sqrt{6})^2 = 3^2 + 11^2 - 88 \sqrt{6} + 96[/imath].

Just a thought... I'm getting the idea that you haven't learned how to solve Quadratic equations, nor how to FOIL out an expression?

-Dan

I understand very well about foil because it's very common but i have problem on quadratic equation.. Would you suggest me something so that i can more understand.... Thank zillion times man...

 

[Murshed]

What is the meaning of [imath]\Delta'[/imath]? Does it perhaps stand for "the value of the discriminant"? If so, then your value of "121" cannot be correct. (It is 22 that is being squared, not 11.) The value of the square root of [imath]\Delta'[/imath] is correct, however.

What were the instructions for this drawing? Are you supposed to be finding the *shaded* area?

Thank you!

You're welcome as always

Actually, i found it somewhere so i became curious to learn that's all about it

 

[Murshed]

b²-4ac = 22² - 4*1*25 = (2*11)² - 4*25 = 4*121-4*25 = 4(121-25) = 4(96). That is why they use Δ′ as Δ/4

This is a perfect case that shows that one needs to define their variables.

Nice explanation sir



Would you suggest any YouTube channel so that i can understand it better in real life example...

 

[Murshed]

I would like to disturb you by sending screenshot.. Thank you sir

 

[Dr.Peterson]

That is why they use Δ′ as Δ/4

This is a perfect case that shows that one needs to define their variables.

Presumably, wherever the solution came from (and we haven't even been shown the full statement of the problem), [imath]\Delta'[/imath] is a standard notation that was previously defined (e.g. in a textbook).

If so, then it's not so much a need for defining variables in writing a solution, as a need to state the meaning of notation when communicating with people elsewhere who may use different notation. Many people don't realize how non-universal math is.