If alpha is a repeated root of multiplicity 2 | vedclass

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Question

(A) Let $f(x) = 18x^3-33x^2+20x-4$. Since $\alpha$ is a repeated root of multiplicity $2$,it must satisfy $f(\alpha) = 0$ and $f'(\alpha) = 0$.First,f

Vedclass · Accepted Answer

$3\alpha^2-8\alpha+4=0$

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(A) Let $f(x) = 18x^3-33x^2+20x-4$. Since $\alpha$ is a repeated root of multiplicity $2$,it must satisfy $f(\alpha) = 0$ and $f'(\alpha) = 0$.First,find the derivative: $f'(x) = 54x^2-66x+20$.Setting $f'(\alpha) = 0$ gives $54\alpha^2-66\alpha+20 = 0$.Dividing the entire equation by $2$,we get $27\alpha^2-33\alpha+10 = 0$.We also know $f(\alpha) = 18\alpha^3-33\alpha^2+20\alpha-4 = 0$.From $f'(\alpha) = 0$,we have $33\alpha^2 = 27\alpha^2+10$,so $33\alpha^2 = 27\alpha^2+10$.Alternatively,solving $27\alpha^2-33\alpha+10 = 0$ using the quadratic formula: $\alpha = \frac{33 \pm \sqrt{1089-1080}}{54} = \frac{33 \pm 3}{54}$.This gives $\alpha = \frac{36}{54} = \frac{2}{3}$ or $\alpha = \frac{30}{54} = \frac{5}{9}$.Testing $\alpha = \frac{2}{3}$ in $f(x)$: $18(\frac{8}{27}) - 33(\frac{4}{9}) + 20(\frac{2}{3}) - 4 = \frac{16}{3} - \frac{44}{3} + \frac{40}{3} - \frac{12}{3} = 0$.Thus,$\alpha = \frac{2}{3}$ is the root.Substituting $\alpha = \frac{2}{3}$ into the options: $3(\frac{2}{3})^2 - 8(\frac{2}{3}) + 4 = 3(\frac{4}{9}) - \frac{16}{3} + 4 = \frac{4}{3} - \frac{16}{3} + \frac{12}{3} = 0$.This matches option $A$.

If $\alpha$ is a repeated root of multiplicity $2$ of the equation $18x^3-33x^2+20x-4=0$,then

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