1. Suppose k is not equal to 0. Show that the quadratic equation kx^2-2x-3k=0 must have real roots.
2. Show that x^2 - x - 4 - 4/x + 16/x^2 = (x + 4/x)^2 - (x + 4/x) - 12
questions on quadratic equations
2007-01-01 4:45 am
回答 (2)
2007-01-01 4:50 am
✔ 最佳答案
1Consider discriminant
discriminant
=(-2)^2-4k(-3k)
=4+12k^2
>0
This show that the quadratic equation kx^2-2x-3k=0 must have real roots.
2
(x + 4/x)^2 - (x + 4/x) - 12
=x^2+8+16/x^2-x-4/x-12
=x^2-x-4-4/x+16/x^2
2007-01-01 5:00 am
1. delta= b^2-4ac
=2^2-4*k*(-3k)
=4+12k^2
as k^2 >0,
delta>0
so it must have real roots
2. R.H.S.=(x + 4/x)^2 - (x + 4/x) - 12
=x^2 + 8 + 16/x^2 - x - 4/x -12
=x^2 - x - 4 - 4/x + 16/x^2 =L.H.S #
=2^2-4*k*(-3k)
=4+12k^2
as k^2 >0,
delta>0
so it must have real roots
2. R.H.S.=(x + 4/x)^2 - (x + 4/x) - 12
=x^2 + 8 + 16/x^2 - x - 4/x -12
=x^2 - x - 4 - 4/x + 16/x^2 =L.H.S #
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