4x²-5=2(x+1)²-7 discuss nature of root
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hey mate ☺️
4x^2-5=2(x+1)^2-7
=>4x^2-5=2x^2+4x+2-7
=>4x^2-2x^2-4x-5+5=0
=>2x^2-4x=0
=>2x^2-4x+0=0
here,a=2,b=(-4),c=0
now,b^-4ac(D)=(-4)^2-4.2.0
=16-0
=16
°•°D>0, roots are real and unequal
❤️HOPE IT WILL HELP YOU❤️
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