Question

06- Find the roots of the equation x3 + 14x2 + 56x - 64 = 0,
roots being in A.P

(A) 2, 4,8
(B) -2,-4,-8
(C) 2, 2, -1
(D) 2, 2, 1
O
A
O
B​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Step-by-step explanation:

Suppose we wish to solve the equation x3 −6x2 +11x−6 =0 This equation can be factorised to give (x −1)(x −2)(x−3) = 0 This equation has three real roots, all different -the solutions are x = 1, x = 2 and x = 3. In Figure 1 we show the graph of y = x3 −6x2 +11x−6. y 3 2 1-1-1-2-3 1 2 3 x Figure 1. The graph of y = x3 −6x2 +11x−6. Notice that it starts low down on the left, because as x gets large and negative so does x3 and it finishes higher to the right because as x gets large and positive so does x3. The curve crosses the x-axis three times, once where x = 1, once where x = 2 and once where x = 3. This gives us our three separate solutions.