find all the zeroes of the quadratic poly. abx^2+(b^2-a)

Question

anonymous · Answer

@apoorvk

anonymous · Answer

@Diyadiya

apoorvk · Answer

So equate this thing to zero.

$$abx^2 + (b^2-a) = 0$$$$or,~ x^2 = \frac{a - b^2}{ab}$$$$or, x = \pm\sqrt{\frac{a - b^2}{ab}}$$



Does this help?

anonymous · Answer

can u make it into de stndrd form,,,, ax^2+bx+c=0

apoorvk · Answer

Oh yeah, we can work that way too..

comparing coefficients with std. form $Ax^2 + Bx + C=0$

what would A, B and C be according to you?

anonymous · Answer

i guess a=ab, b=0, c=b^2-a??

apoorvk · Answer

yes!!

So, 

use this is in the quadratic formula:
$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$

apoorvk · Answer

Since B = 0, you will get the same thing basically.

So, this quadratic formula process is similar to taking the route around the globe! o.O

anonymous · Answer

itz damn so confusing...let ma teachr suffr de rrest....helpin me teach dis ;)

apoorvk · Answer

since B=0, $\large  x = \frac{\sqrt{-4AC}}{2A} = \sqrt{\frac {-4AC}{4A^2}} = \sqrt{\frac {-C}{A}}$



Does this make it simple?

anonymous · Answer

ummm....nice try... bt ma head is kindah dumb to undrstnd dis....itz k jz leav it...