Math, asked by khushi3190, 1 year ago

slove with quadratic formula plzzz...

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Answered by Akshta1

ur ANSWER IN THE ABOVE PIC ..

PLSS MRK BRAINLIEST

Good luck for exam ..!!!

I am also going to give my maths board exam today...

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Akshta1: in second pic applied quadratic formula

Akshta1: kindly first see the content. properly amd then raise ur questions...

aditykhebade2012: Sorry for the late reply but,

aditykhebade2012: In the 4th line(where D=4-4[1^2-(p^2)^2] you later on simplify it indo D=4[1-1+p^4] . WHere did the other 4 go?

aditykhebade2012: PLease do correct me if im wrong:

aditykhebade2012: Correction to your answer would be: D=4-4[(1+p^2)(1-p^2)

aditykhebade2012: D=4-4[1(1-p^2)+p^2(1-p^2)] D=4-4[1-p^2+p^2-p^4

aditykhebade2012: D=4-4[1-p^2+p^2-p^4

aditykhebade2012: the overall what im trying to say is that its should be -4p^4

khushi3190: how was ur exam

Answered by aditykhebade2012

Answer:

Step-by-step explanation:

(1+p²)x²+2x+(1-p²)=0

By quadratic formula

a=1+p² ;b =2 ; c= 1-p²

x= $\frac{-b±\sqrt{b^{2}-4ac } }{2a}$ [IGNORE the A thingy]

x= $\frac{-2±√2x^{2}-4×1+p²×1-p²}{2×1+p²}$

x= $\frac{-2±√2x²-4×(1+p²)×(1-p²)}{2+2p²}$

x=-2±√4×1+p²×1-p²4-4p²+4p²-4p^4/2+2p^2

x=2±√4×1+p²×1-p²4-4p^4/2+2p^2

x= $\frac{2+2-2p^{2} }{2+4p^{2} }$ OR x= $\frac{2-2-2p^{2} }{2+4p^{2} }$

x= $\frac{4-2p^{2} }{2+2p^{2} }$ OR x= $\frac{0-2p^{2} }{2+2p^{2} }$

x= $\frac{2(2-p^{2}) }{2(1+p^{2}) }$ OR x= $\frac{0-2p^{2} }{2(+p^{2}) }$

x= $\frac{(2-p^{2}) }{(1+p^{2}) }$ OR x= $\frac{0-^{2} }{2(+p^{2}) }$

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