Question

Don't Delete this question considering to be plagiarised.
Previous question was wrong so had to post it again.

$\sf If \: the \: equation \:4x^3+20x^2-23x+6=0$
$\sf \: have \: two \: of \: the \: roots \: \alpha \: \sf and \: \beta \:equal\: ,then$
$\sf \: \boxed{\sf\red{\gamma + { |\frac{a}{b}| }}}$ $\sf \: will \: be -$

$\bf Note- Consider,\: \gamma\: as \:third \: root$​

Answer 1

Appropriate Question :-

If the equation 4x³ + 20x² - 23x + 6 = 0 have two of the roots $\alpha$ and $\beta$ equal, then $\gamma + \frac{\alpha }{\beta}$ will be :-

Answer :-

• -5

Step by step explanation :-

Here,

According to the question,

$\rm\implies{ \alpha = \beta - - - - (1) }$

Let,

The roots of the equation be $\alpha , \beta , \gamma$ .

Given equation : 4x³ + 20x² - 23x + 6 = 0

For making the answer easy we have to take $\sum$ of x.

After solving we get,

$\sf\implies{ \sum x = \alpha + \beta + \gamma = - \frac{20}{4} }$

From equation (1) we can be write,

$\sf\implies{ \sum x = \alpha + \alpha + \gamma = \frac{ - 20}{4} = - 5 }$

Also can be written as,

$\sf\implies{2 \alpha + \beta = - 5} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{ \beta = - 5 - 2 \alpha - - - - (2)}$

Now taking $\sum$ of xy.

$\sf\implies{ \sum xy = \alpha ( \alpha ) + \alpha ( \beta ) + \beta ( \alpha ) = \frac{ - 23}{4} }$

Now, it becomes

$\sf\implies{ \sum xy = \alpha {}^{2} + 2 \alpha \beta = \frac{ - 23}{4} }$

Transporting 4 to LHS from RHS, we get

$\sf\implies{4( \alpha {}^{2} + 2 \alpha \beta) = - 23 } \: \\ \\ \sf\implies{4 \alpha {}^{2} + 8 \alpha \beta + 23 = 0}$

Substituting equation (2),

$\sf\implies{ 4\alpha {}^{2} + 8 \alpha( - 5 - 2 \alpha ) + 23 = 0 } \\ \\ \sf\implies{4 \alpha {}^{2} - 40 \alpha - 16 \alpha {}^{2} + 23 = 0 } \: \: \: \\ \\ \sf\implies{12 \alpha {}^{2} + 40 \alpha - 23 = 0 } \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Using factorisation method,

$\sf\implies{12 \alpha {}^{2} - 6 \alpha + 46 \alpha - 23 = 0 } \\ \\ \sf\implies{ \alpha = \frac{1}{2} ,- \frac{23}{6} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Substituting $\alpha$ in equation (2),

$\sf\implies{ \beta = - 5 - 2 \bigg\lgroup \frac{1}{2} \bigg\rgroup} \\ \\ \sf\implies{ \beta = - 5 - 1} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\sf\implies \red{ \beta = - 6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf\implies{ \beta = - 5 - 2 \bigg( - \frac{23}{6} \bigg)} \\ \\ \sf\implies \red{ \beta = - \frac{8}{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Taking $\sum$ of xyz.

$\sf\implies{ \sum xyz = \alpha {}^{2} \beta = - \frac{6}{4} = - \frac{3}{2} }$

Product of roots do not satisfy $\alpha = - \frac{8}{3}$

So roots are,

$\sf\implies{ \frac{1}{2} ,\frac{1}{2}, - 6 }$

Verification :-

From equation (1)

$\sf \implies{ \alpha = \beta } \: \: \\ \\ \sf \implies{ \frac{1}{2} = \frac{1}{2} }$

$\sf \implies{ \gamma = - 6}$

We have to find,

$\sf \implies{ \gamma + \frac{ \alpha }{ \beta } = - 6 + \frac{ \large\frac{1}{2} }{ \large \frac{1}{2} } } \\ \\ \sf \implies{ - 6 + 1 = \boxed{ - 5}} \: \: \: \: \:$

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