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Mathematics 20 Online
OpenStudy (anonymous):

Quadratic Equation: (x-4)(x-6)=2

terenzreignz (terenzreignz):

You're supposed to find x, am I right? :)

OpenStudy (anonymous):

haha i dont know :) i think yes?

terenzreignz (terenzreignz):

Well, perform the FOIL method on (x-4)(x-6). You get...?

OpenStudy (anonymous):

what will you do with 2?

terenzreignz (terenzreignz):

Patience :) FOIL method on the left-hand-side you get x² - 10x + 24 right?

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

so... x² - 10x + 24 = 2 Now you can more clearly see what to do with the 2, am I right? ;)

OpenStudy (anonymous):

.....

OpenStudy (anonymous):

you cant find x

OpenStudy (anonymous):

then what is it?

OpenStudy (anonymous):

use FOIL

OpenStudy (anonymous):

know what that is?

OpenStudy (anonymous):

yes. i got x^2-10x+24=2

OpenStudy (anonymous):

then whats next?

terenzreignz (terenzreignz):

Now add -2 to both sides

OpenStudy (anonymous):

what?

terenzreignz (terenzreignz):

Well, transpose 2, from the right side to the left... x² - 10x + 22 = 0

OpenStudy (anonymous):

oohh i get it. then what?

terenzreignz (terenzreignz):

Use the quadratic formula. Everything falls into place.

OpenStudy (anonymous):

i really cant. please help me :(

terenzreignz (terenzreignz):

Given your equation in the form ax² + bx + c = 0 Your x values are given by \[\huge \frac{-b \ \pm \ \sqrt{b^2-4ac}}{2a}\]

terenzreignz (terenzreignz):

Plug in and solve.

OpenStudy (anonymous):

what is my a ? my b is 10? my c is 24 ?

OpenStudy (anonymous):

i mean my c is 20?

terenzreignz (terenzreignz):

your b is -10 not just 10 a is 1, as there is no coefficient.

terenzreignz (terenzreignz):

c is 22, not 24. Look again.

OpenStudy (anonymous):

oh okay :) then whats next?

OpenStudy (anonymous):

ill just solve it?

terenzreignz (terenzreignz):

Yep.

OpenStudy (anonymous):

i think we're going to do a standard equation not quadratic equation?!

OpenStudy (anonymous):

standard form rather

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