0,1,-3÷2-5÷2 in theory of eqution
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1
3x
3
–x+88=4x
2
if one of the root of the given cubic polynomial equation is \mathbf{2\;-\;\sqrt{7}\;i}2−
7
i.
Solution:
Let, f (x) = 3x^{3} – x + 88 – 4x^{2}3x
3
–x+88–4x
2
The given cubic function will have three roots i.e. α, β, and γ.
Therefore, α = x = \mathbf{2\;-\;\sqrt{7}\;i}2−
7
i
And, β = x = \mathbf{2\;+\;\sqrt{7}\;i}2+
7
i [Since, imaginary roots occur in conjugate pairs]
Therefore, \mathbf{(x\;-\;2\;+\;\sqrt{7}\;i)\;(x\;-\;2\;-\;\sqrt{7}\;i)}(x−2+
7
i)(x−2−
7
i) = (x – 2)^{2} + 7 = x^{2} – 4x + 11(x–2)
2
+7=x
2
–4x+11 = g (x)
On dividing f (x) by g (x) we will get 3x + 8 as quotient.
Therefore, 3x + 8 = 0
Hence, \mathbf{\gamma \;=\;-\;\frac{8}{3}}γ=−
3
8
Therefore, the roots of given
Equation f (x) are \mathbf{2\;-\;\sqrt{7}\;i}2−
7
i, \mathbf{2\;+\;\sqrt{7}\;i}2+
7
i, \mathbf{\gamma \;=\;-\;\frac{8}{3}}γ=−
3
8
⇒ If any given equation f (x) of nth degree has maximum ‘p’ positive real roots and ‘q’ negative real roots, then the given equation has at least n – (p + q) imaginary roots.
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