is someone to help me. solve the below question from ch 4 quadratic equations.. Created by sindhusharavuri. Math

is someone to help me. solve the below question from ch 4 quadratic equations...

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sindhusharavuri @sindhusharavuri

July 2021 2 60 Report

is someone to help me. solve the below question from ch 4 quadratic equations.

DhanyaDA

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Hi friend!!!!

___________________________

Given p(x)=x²+2x+4=0

=x²+2x+1+3=0

=(x+1)²+3=0

=(x+1)²=-3

=x+1=±√[-3]

x=±√(-3)-1

__________________________

I hope this will help u ;)

pankaj12je

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Hey there !!!

x²+2x+4=0----------------Equation 1

x²+2x+1+3=0

(x+1)²+3=0

(x+1)²=-3

x+1=√-3

x=-1+√-3

Value of x is imaginary as discriminant i.e, b²-4ac < 0

2²-4*1*4 <0

So value of x= √-3-1.

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