Question

If α and ß are the roots of the quadratic equation 3x^2 - 5x + 4 = 0 , find the value of 1/α + 1/ß - 2αß.

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Please, Solve it
25 - points​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{-\frac{17} {12} }}}$

$\rule{200}{4}$

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

As we know,

$\text{In the equation\ ax^2+bx+c=0, if \ x\in\{\alpha,\ \beta \}, }\\\\\text{then \ \alpha+\beta=-\dfrac{b}{a}\quad\&\quad\alpha\beta=\dfrac{c}{a}. }$

So,

$a=3\quad;\quad b=-5\quad;\quad c=4\\\\\\\alpha+\beta=-\dfrac{-5}{3}=\dfrac{5}{3}\\\\\\\alpha\beta=\dfrac{4}{3}$

Now,

$\begin{aligned}&\dfrac{1}{\alpha}+\dfrac{1}{\beta}-2\alpha\beta\\\\\implies\ \ &\dfrac{\alpha+\beta}{\alpha\beta}-2\alpha\beta\\\\\implies\ \ &\dfrac{\left(\dfrac{5}{3}\right)}{\left(\dfrac{4}{3}\right)}-2\cdot\dfrac{4}{3}\\\\\implies\ \ &\dfrac{5}{4}-\dfrac{8}{3}\\\\\implies\ \ &\mathbf{-\dfrac{17}{12}}\end{aligned}$

Answer 2

Answer:

${\bold{\therefore \frac{1}{\alpha}+\frac{1}{\beta}-2\alpha \beta=\frac{-17}{12}}}$

Step-by-step explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a quadratic eqn and its roots are given.

• We have to find the given value.

$\underline \bold{Given : } \\ \implies {3x}^{2} - 5x + 4 = 0 \\ \implies Roots = \alpha \: and \: \beta \\ \\ \underline \bold{To \: Find : } \\ \implies \frac{1}{ \alpha } + \frac{1}{ \beta } - { 2\alpha \beta } = ?$

• According to given question :

$\implies a = 3 \\ \implies b = - 5 \\ \implies c = 4 \\ \\ \bold{first \: zeroes} \\ \implies \alpha + \beta = \frac{ - b}{a} \\ \implies \alpha + \beta = \frac{ - ( - 5)}{3} = \frac{5}{3} \\ \\ \bold {second \: zeroes } \\ \implies \alpha \beta = \frac{c}{a} \\ \implies \alpha \beta = \frac{4}{3} \\ \\ \implies \frac{1}{ \alpha } + \frac{1}{ \beta } - 2 \alpha \beta \\ \implies \frac{ \beta + \alpha }{ \alpha \beta } - 2 \alpha \beta \\ \implies \frac{ \alpha + \beta }{ \alpha \beta } - 2 \alpha \beta \\ \implies \frac{ \frac{5}{3} }{ \frac{4}{3} } - 2 \times \frac{4}{3} \\ \implies \frac{5}{3} \times \frac{3}{4} - \frac{8}{3} \\ \implies \frac{5}{4} - \frac{8}{3} \\ \implies \frac{15 - 32}{12} \\ \bold{\implies \frac{ - 17}{12} } \\ \\ \bold{\therefore \frac{1}{ \alpha } + \frac{1}{ \beta } - 2 \alpha \beta = \frac{ - 17}{12} }$