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# Nature of roots

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Determine the nature of the roots without solving the equation.

\(kx^2+(4k-1)x=2-4k\)

I did this....

\(kx^2+4kx-x+4k-2=0 \)

\(kx^2+x(4k-1)+2(2k-1)=0\)

Not sure..what I would not be suprised of is if I had made a mistake......

Thank you for all your help..always appreciated...

Mar 3, 2021

#1
+802
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But then I did this: leave it as

\(kx^2+x(4k-1)+4k-2\)

\(d=b^2-4ac\)

\(=(4k-1)^2-4(k)(4k-2)\)

\(=16k^2-8k+1-16k^2+8k\)

\(=1\)

So the roots are Real, un-equal and rational....is this right perhaps?

Mar 3, 2021
#2
+112972
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Yes, that is correct.

The symbol for the determinant is a triangle.

Backslash triangle.

\(\triangle \)

Melody  Mar 3, 2021
#3
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Hello dear Melody.....

Thank you for confirming....yes I know its the triangle, just do not know where to get the symbol...I searched but could not find it...

juriemagic  Mar 4, 2021
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+112972
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