a Amaths. Quadratic Equations question
2006-11-16 8:01 am
Show that the quardratic equation 2x^2 - 3(m-1)x + (m^2-4m-6) = 0 has two unequal real roots in x.
回答 (2)
2006-11-16 8:14 am
For the equation 2x^2 - 3(m-1)x + (m^2-4m-6) = 0
Discriminant
= [-3 (m-1)]^2 - 4 (2) (m^2 - 4m - 6)
= 9 (m - 1)^2 - 8 (m^2 - 4m - 6)
= 9 (m^2 - 2m + 1) - 8m^2 + 32m + 48
= 9m^2 - 18m + 9 - 8m^2 + 32m + 48
= m^2 + 14m + 57
= m^2 + 2(7)m + (7)^2 -(7)^2 + 57
= (m + 7)^2 - 49 + 57
= (m + 7)^2 + 8
> 0
Hence the qiven equation has 2 distinct real roots
Discriminant
= [-3 (m-1)]^2 - 4 (2) (m^2 - 4m - 6)
= 9 (m - 1)^2 - 8 (m^2 - 4m - 6)
= 9 (m^2 - 2m + 1) - 8m^2 + 32m + 48
= 9m^2 - 18m + 9 - 8m^2 + 32m + 48
= m^2 + 14m + 57
= m^2 + 2(7)m + (7)^2 -(7)^2 + 57
= (m + 7)^2 - 49 + 57
= (m + 7)^2 + 8
> 0
Hence the qiven equation has 2 distinct real roots
參考: me
2006-11-16 8:14 am
D = 9(m-1)^2 - 4*2*(m^2-4m-6)
=m^2 + 14m + 57
=(m^2 + 14m + 49) + 57 - 49
=(m + 7)^2 + 8
Since (m+7)^2 is always greater than or equal to 0 for all real values of m,
D > 0.
So the equation has two unequal real roots.
=m^2 + 14m + 57
=(m^2 + 14m + 49) + 57 - 49
=(m + 7)^2 + 8
Since (m+7)^2 is always greater than or equal to 0 for all real values of m,
D > 0.
So the equation has two unequal real roots.
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