bhuvaneshnick
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- Dec 18, 2014
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Example 1.6 Solve the equation x4 - 2x3 - 21x2 + 22x + 40 = 0 whose roots are in A.P.
Solution Let the roots be a - 3d, a - d, a + d, and a + 3d, so the roots equal 4a = 2. Then a = 2.
Also, the product of the roots is (a2 - 9d 2)(a2 - d 2) = 40, so:
. . . . .\(\displaystyle \left(\dfrac{1}{4}\, -\, 9d^2\right)\left(\dfrac{1}{4}\, -\, d^2\right)\, =\, 40\) or \(\displaystyle 144d^4\, -\, 40d^2 \, -\, 639\, =\, 0\)
Then d 2 = 9/4 or -7/36. Thus, \(\displaystyle d\, =\, \pm 3/2;\) the other value is inadmissible. Hence, the required roots are -4, -1, 2, 5.
my question: how the addition of geometric series gives 2 and product gives 40
Solution Let the roots be a - 3d, a - d, a + d, and a + 3d, so the roots equal 4a = 2. Then a = 2.
Also, the product of the roots is (a2 - 9d 2)(a2 - d 2) = 40, so:
. . . . .\(\displaystyle \left(\dfrac{1}{4}\, -\, 9d^2\right)\left(\dfrac{1}{4}\, -\, d^2\right)\, =\, 40\) or \(\displaystyle 144d^4\, -\, 40d^2 \, -\, 639\, =\, 0\)
Then d 2 = 9/4 or -7/36. Thus, \(\displaystyle d\, =\, \pm 3/2;\) the other value is inadmissible. Hence, the required roots are -4, -1, 2, 5.
my question: how the addition of geometric series gives 2 and product gives 40
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