If the graph y=3kx^2+(k+9)x+4 touches the x-axis at one point. Find k.
Which of the following equation(s) has/have equal real roots?
I) x^2=2x+1
II) x^2=2x-1
III) x^2=0
Maths (With steps)
2008-10-22 1:51 am
回答 (2)
2008-10-22 2:17 am
✔ 最佳答案
1. (k+9)^2-4(3k)(4)=0k^2+18k+81-48k=0
k^2-30k+81=0
(k-27)(k-3)=0
then k=27 or k=3
2. x^2=2x+1
x^2-2x-1=0
(cross method)
no equal real roots.
x^2=2x-1
x^2-2x+1=0
(x-1)(x-1)=0
then, x=1
so, this equation has equal real roots.
x^2=0
(x-0)(x-0)=0
then, x=0
so, this equation has equal real roots.
2008-10-22 2:21 am
If the graph touches the x-axis at one point
Thenthe equation 3kx^2+(k+9)x+4 = 0 has equal roots
△ = (k+9)^2 - 4(3k)(4) = 0
k^2 + 18k + 81 - 48k = 0
k^2 - 30k + 81 = 0
(k-27)(k-3) = 0
k = 3 or 27
I) x^2 = 2x + 1
x^2 - 2x - 1 = 0
△ = (-2)^2 - 4(1)(-1) = 8 (≠0)
II) x^2=2x-1
x^2 - 2x + 1 = 0
△ = (-2)^2 - 4(1)(1) = 0
III) x^2 = 0
△ = 0^2 - 4(1)(0) = 0
∴ (II) and (III) has equal real roots
Thenthe equation 3kx^2+(k+9)x+4 = 0 has equal roots
△ = (k+9)^2 - 4(3k)(4) = 0
k^2 + 18k + 81 - 48k = 0
k^2 - 30k + 81 = 0
(k-27)(k-3) = 0
k = 3 or 27
I) x^2 = 2x + 1
x^2 - 2x - 1 = 0
△ = (-2)^2 - 4(1)(-1) = 8 (≠0)
II) x^2=2x-1
x^2 - 2x + 1 = 0
△ = (-2)^2 - 4(1)(1) = 0
III) x^2 = 0
△ = 0^2 - 4(1)(0) = 0
∴ (II) and (III) has equal real roots
收錄日期: 2021-04-20 21:34:04
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