Question

If a and ß are the zeroes of the quadratic polynomial 3x^2 + 7x - 20, then find the value of 1/alpha^2 + 1/beta^2

Answer 1

Answer:

answer is 85

step by step :

alpha +be a = 7

alphabeta = -18

alpha2 + beta2

(alpha+beta)2 -2*alpha*beta

= (7)2-2(-18)

= 85

Answer 2

⠀⠀⠀$\huge\underline{ \mathrm{ \red{QueS{\pink{tiOn}}}}}$

If a and ß are the zeroes of the quadratic polynomial 3x^2 + 7x - 20, then find the value of 1/alpha^2 + 1/beta^2

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⠀⠀⠀$\huge{ \underline{ \purple{ \bold{ \underline{ \mathrm{ExPlanA{\green{TiOn }}}}}}}}$

$\large\underline{ \underline{ \red{ \bold {Given}}}}$

$\bf\:p(x) = {3x}^{2} + 7x - 20$

$\large\underline{ \underline{ \red{ \bold {To \:Find}}}}$

$\bf\color{purple}{we \: need \: to \: find \: the \: value \: of \: \frac{1}{ \alpha {}^{2} } + \frac{1}{ { \beta }^{2} } }$

⠀⠀⠀⠀⠀$\huge\underline{ \underline{ \orange{ \bold{sOluTiOn}}}}$

⠀⠀⠀⠀⠀

$\bf\: firstly\: we \:find \:the\: zeroes \:of\:{ 3x}^{2}+7x-20$

$\color{blue}{: \implies} \bf{3 {x}^{2} + 7x - 20 }$

$\color{blue}{: \implies} \bf{3 {x}^{2} + 12x - 5x - 20} \\ \\ \color{blue}{: \implies} \bf \: 3x(x + 4) - 5(x + 4) \\ \\\color{blue}{: \implies} \bf \: (x + 4)(3x - 5) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{\color{blue} \bf x = - 4 \: \: \: or \: \: \: x = \frac{5}{3}}}$

Now finding the value of $\bf\:\frac{1}{\alpha{}^{2}}+\frac{1}{\beta{}^{2}}$

$\bf\: { \color{green}: \implies{ \alpha = - 4 \: \: and \: \: \beta = \frac{5}{3} }} \\ \\ \bf\: { \color{green}: \implies{ \frac{1 }{ \alpha {}^{2} } + \frac{1}{ \beta {}^{2} } }}$

$\bf\: { \color{green}: \implies{ \frac{1}{ - 4 {}^{2} + } + \frac{ \frac{1}{5 {}^{2} } }{3 {}^{2} } }} \\ \\\bf\: { \color{green}: \implies{ \frac{1}{16} + \frac{9}{25} }} \\ \\\bf\: { \color{green}: \implies{ \frac{25 + 144}{400} }} \\ \\ \bf\: { \color{green}: \implies{ \frac{169}{400} }}$

$\\ \\\bf\: { \color{green}: \implies{ \underline{ \boxed{ \color{red}{{ \frac{169}{400} }} }}}}$

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