is someone to help me. solve the below question from ch 4 quadratic equations.
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Answered by
3
Hi friend!!!!
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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 ;)
___________________________
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 ;)
Answered by
2
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.
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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