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- I. x^3 – 9x^2 + 8x = 0 II. y^3 + 7y^2 + 12y = 0
Question 1:
I. x^3 – 9x^2 + 8x = 0 II. y^3 + 7y^2 + 12y = 0
A. if x > y
B. if x ≤ y
C. if x ≥ y
Feedback According to the given equations :
I. x^3 – 9x^2 + 8x = 0
(x^3 – 9x^2 + 8x)/x = 0/x
x2 – 9x + 8 = 0
x2 – 8x – x + 8 = 0
x (x – 8) –1(x – 8) = 0
(x – 1) (x – 8) = 0
x = 0, 1, 8
II. y^3 + 7y^2 + 12y = 0
(y^3 + 7y^2 + 12y)/y = 0/y
y2 + 7y + 12 = 0
y^2 + 4y + 3y + 12 = 0
y(y + 4) + 3 (y + 4) = 0
(y + 3) (y + 4) = 0y = 0, – 3, – 4
Therefore, x ≥ y
Hence, option C is correct.
I. x^3 – 9x^2 + 8x = 0
(x^3 – 9x^2 + 8x)/x = 0/x
x2 – 9x + 8 = 0
x2 – 8x – x + 8 = 0
x (x – 8) –1(x – 8) = 0
(x – 1) (x – 8) = 0
x = 0, 1, 8
II. y^3 + 7y^2 + 12y = 0
(y^3 + 7y^2 + 12y)/y = 0/y
y2 + 7y + 12 = 0
y^2 + 4y + 3y + 12 = 0
y(y + 4) + 3 (y + 4) = 0
(y + 3) (y + 4) = 0y = 0, – 3, – 4
Therefore, x ≥ y
Hence, option C is correct.
D. if x < y
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