Question

Find the value of $4\beta -a\alpha $ If $a>b>c$ and ${{a}^{3}}+{{b}^{3}}+27ab\ =729$ then the quadratic equation $a{{x}^{2}}+bx-9=\ 0$ has roots $\alpha, \beta \left( \alpha <\beta \right)$.

Answer 1

Hint: Relation between roots of a quadratic equation $\text{A}{{x}^{2}}+\text{B}x+\text{c}=\ 0$ is given as
Sum of roots $=\ \dfrac{-\text{B}}{\text{A}}$ and Product of roots $=\ \dfrac{\text{C}}{\text{A}}$
Use these properties to get equations which have relations among $\alpha ,\beta $, a and b. Now, find the values of ‘a’ and ‘b’ in terms of ‘$\alpha $’ and ‘$\beta $’ and put them in the given equation ${{a}^{3}}+{{b}^{3}}+27ab\ =729$. Now, evaluate the value of the given expression and use the given relations in (a, b) and ($\alpha ,\beta $) i.e. $a>b>c$ and $\left( \alpha <\beta \right)$.

Complete Step-by-Step solution:

Here, it is given that $a>b>c$ and ${{a}^{3}}+{{b}^{3}}+27ab\ =729$, then we need to determine the value of $a{{x}^{2}}+bx-9=\ 0$ are ($\alpha ,\beta $) with the condition $\left( \alpha <\beta \right)$.
So, we have
${{a}^{3}}+{{b}^{3}}+27ab\ =729$ …………………………………………………….(i)
And we have a quadratic equation given below with roots $\alpha $ and $\beta $ as
$a{{x}^{2}}+bx-9=\ 0$ ……………………………………………………………(ii)
Now, we know the relation between the roots and coefficients of any quadratic $\text{A}{{x}^{2}}+\text{B}x+\text{c}=\ 0$ is given as
Sum of roots $=\ \dfrac{-\text{B}}{\text{A}}$ ……………………………………………………………(iii)
Product of roots $=\ \dfrac{\text{C}}{\text{A}}$ …………………………………………………………..(iv)
Hence, we can write the two equations from the equation (ii) using the relation expressed in equations (iii) and (iv). So, we get
$\alpha +\beta \ =\ \ \dfrac{-b}{a}$ and $\alpha \beta \ =\ \ \dfrac{-9}{a}$ …………………………………………………….(v)
So, we can get values of a and b in terms of ‘$\alpha $’ and ‘$\beta $’ by solving the equations given in equation (v).
So, we get
$\dfrac{-b}{a}=\ \alpha +\beta $ and $\dfrac{-9}{a}\ =\ \alpha \beta \ $
So, we get values of ‘a’ from the second relation mentioned just above as
$a\ =\ \ \dfrac{-9}{\alpha \beta }$ ………………………………………………………………………………………..(vi)
Now, put value of ‘a’ in the first relation of equation (v), so, we get
$\dfrac{-b}{\dfrac{-9}{\alpha \beta }}=\ \alpha +\beta $
 $-b\ =\ \ \left( \alpha +\beta \right)\left( \dfrac{-9}{\alpha \beta } \right)$
$b\ =\ \ \dfrac{9\left( \alpha +\beta \right)}{\alpha \beta }$ …………………………………………………………………………………….(vii)
Now, we can put values of ‘a’ and ‘b’ from the equation (vi) and (vii) to the equation (i). So, we get
${{\left( \dfrac{-9}{\alpha \beta } \right)}^{3}}+{{\left( \dfrac{9\left( \alpha +\beta \right)}{\alpha \beta } \right)}^{3}}+27\left( \dfrac{-9}{\alpha \beta } \right)\times \left( \dfrac{9\left( \alpha +\beta \right)}{\left( \alpha \beta \right)} \right)\ =\ 729$
Now we know ${{9}^{3}}\ =\ 729$ and hence, simplifying the above equation we get
$\dfrac{-729}{{{\left( \alpha \beta \right)}^{3}}}+\dfrac{729{{\left( \alpha +\beta \right)}^{3}}}{{{\left( \alpha \beta \right)}^{3}}}-\left( \dfrac{729\times 3\left( \alpha +\beta \right)}{{{\left( \alpha \beta \right)}^{2}}} \right)\ =\ 729$
Now, cancelling the term 729 from both sides of the above equation, we get
$\dfrac{-1}{{{\left( \alpha \beta \right)}^{3}}}+\dfrac{{{\left( \alpha +\beta \right)}^{3}}}{{{\left( \alpha \beta \right)}^{3}}}-\dfrac{3\left( \alpha +\beta \right)}{{{\left( \alpha \beta \right)}^{2}}}\ =\ 1$
Now, taking L.C.M. of denominators, we get expression as
$\dfrac{-1+{{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right)}{{{\left( \alpha \beta \right)}^{3}}}=\ \dfrac{1}{1}$
On cross-multiplying the above equation, we get
$-1+{{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right)=\ {{\left( \alpha \beta \right)}^{3}}$ ……………………………………………………..(vii)
Now, we know the algebraic identity of ${{\left( a+b \right)}^{3}}$, which can be given by the following relation as
${{\left( a+b \right)}^{3}}\ =\ {{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ …………………………………………………………………(viii)
Hence, we can replace ${{\left( \alpha +\beta \right)}^{3}}$ using the above identity and hence simplify the equation (vii) as
$-1+{{\alpha }^{3}}+{{\beta }^{3}}+3\alpha \beta \left( \alpha +\beta \right)-3\alpha \beta \left( \alpha +\beta \right)=\ {{\left( \alpha \beta \right)}^{3}}$
$\Rightarrow -1+{{\alpha }^{3}}+{{\beta }^{3}}=\ {{\left( \alpha \beta \right)}^{3}}$
$\Rightarrow {{\alpha }^{3}}+{{\beta }^{3}}-1-{{\left( \alpha \beta \right)}^{3}}=\ 0\ $
\[\Rightarrow \left( {{\alpha }^{3}}-1 \right)+\left( {{\beta }^{3}}-{{\alpha }^{3}}{{\beta }^{3}} \right)=\ 0\ \]
Now, take 1 as common from first two terms and \[{{\beta }^{3}}\] from the last two terms, so we get
\[1\left( {{\alpha }^{3}}-1 \right)+{{\beta }^{3}}\left( 1-{{\alpha }^{3}} \right)=\ 0\ \]
\[\Rightarrow 1\left( {{\alpha }^{3}}-1 \right)+{{\beta }^{3}}\left( {{\alpha }^{3}}-1 \right)=\ 0\ \]
\[\left( {{\alpha }^{3}}-1 \right)\left( 1-{{\beta }^{3}} \right)=\ 0\]
Hence, we know that if \[xy=\ 0\], then x and y should be 0 or both. It means we get
\[{{\alpha }^{3}}-1=\ 0\] or \[1-{{\beta }^{3}}=\ 0\]
\[{{\alpha }^{3}}\ =\ 1\] or \[{{\beta }^{3}}=\ 1\] …………………………………….(viii)
As, we know roots of equation \[{{x}^{3}}\ =\ 1\] are given as
\[{{x}^{3}}\ =\ 1\]
\[\Rightarrow {{x}^{3}}-{{1}^{3}}\ =\ 0\]
Apply the identity of \[\left( {{a}^{3}}\ -{{b}^{3}} \right)\] in the above solution, so, we know
\[\ {{a}^{3}}\ -{{b}^{3}}=\ \left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]
Hence, \[{{x}^{3}}-1\] can be written as
\[\left( x-1 \right)\left( {{x}^{2}}+1+x \right)\ =\ 0\]
So, we get
\[x-1\ =\ 0\] or \[{{x}^{2}}+1+x\ =\ 0\]
\[x=1\] or \[{{x}^{2}}+1+x\ =\ 0\]
So, we get one root of the equation \[{{x}^{3}}-1\] as 1. And we can get the other two roots from the equation \[{{x}^{2}}+1+x\ =\ 0\]. We know the discriminant decides the roots of the quadratic are real or imaginary which is given as
\[\text{D}\ \text{=}\ {{b}^{2}}-4ac\] for quadratic equation \[a{{x}^{2}}+bx+c\ =\ 0\]
So, if D > 0 roots are real and distinct
D < 0 roots are imaginary
D = 0 roots are equal and real
So, discriminant of \[{{x}^{2}}+1+x\ =\ 0\] can be given as
\[\text{D}\ \text{=}\ {{b}^{2}}-4ac\ =\ {{\left( 1 \right)}^{2}}-4\times 1\times 1\ =\ 1-4\ =-3\]
So, D < 0, hence the other two roots of \[{{x}^{3}}-1\] are imaginary.
So, the real root of \[{{\alpha }^{3}}=1\] is 1 and \[{{\beta }^{3}}=1\]is ‘1’ as well. As \[\alpha \]and \[\beta \] can not be imaginary and will be different as it is given \[\left( \alpha <\beta \right)\]. So, values of \[\alpha \]and \[\beta \] will be different and real. So, we can get either of the \[\alpha \]or \[\beta \] as 1 i.e. only one of them will take value as 1.
So, let us suppose \[\beta \ =\ 1\].
Now we can get the value of \[\alpha \] from the equation (vi) and put the value of \[\beta \] equal to 1.
So, we get
\[a\ =\ \dfrac{-9}{\alpha \beta }\ =\ \dfrac{-9}{\alpha }\]
\[\Rightarrow a\alpha \ =\ -9\] …………………………………………………..(ix)
Now, coming to the question point, as we need to determine value of \[4\beta -a\alpha \]; so, we can put value of \[\beta \ =\ 1\] and value of \[a\alpha \ =\ -9\] form equation (ix). So, we get
\[4\beta -a\alpha \ =\ 4\left( 1 \right)-\left( -9 \right)\ =\ 4+9\ =\ 13\]
Hence, 13 is the correct answer.

Note: Another approach for the given question would be that we can factorize the given relation ${{a}^{3}}+{{b}^{3}}+27ab\ =729$ by using the identity
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\ =\ \left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab+bc-ca \right)$
So, we have
${{a}^{3}}+{{b}^{3}}+{{\left( -9 \right)}^{3}}-3\times \left( a \right)\left( b \right)\left( -9 \right)\ =\ 0$
Hence, we get
$\left( a+b-9 \right)\left( {{a}^{2}}+{{b}^{2}}+81-ab+9b-9a \right)\ =\ 0$
As we know the relation
$\dfrac{1}{2}\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right]\ =\ \left( a+b-9 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab+bc-ca \right)$
So, we get
$\dfrac{1}{2}\left( a+b-9 \right)\left( 2{{a}^{2}}+2{{b}^{2}}+162-2ab+18b+18a \right)\ =\ 0$
$\Rightarrow \left( a+b-9 \right)\left( {{\left( a-b \right)}^{2}}+{{\left( b+9 \right)}^{2}}+{{\left( a+9 \right)}^{2}} \right)\ =\ 0$
So, the second bracket will never be ‘0’ as all the terms of them are square of any number and if they will become ‘0’ then we get $a\ =\ b\ =\ -9$ , which is not possible as per the question. So, take $a\ +b\ -9\ =\ 0$ and relate it with the equations of the roots ‘$\alpha $’ and ‘$\beta $’ with the coefficients of the given quadratic equation. So, it can be another approach.
Don’t confuse it with equations \[{{\alpha }^{3}}=1\] or \[{{\beta }^{3}}=1\]. As, we know \[{{x}^{3}}=1\] or \[x={{1}^{\dfrac{1}{3}}}\] will give roots as 1,$w$,${{w}^{2}}$ as \[{{x}^{3}}=1\] is defined by cube of roots of unit. So, if someone knows it then we don’t need to factorize \[{{x}^{3}}-1=\ 0\]. Here, $w\ =\ \dfrac{-1+\sqrt{3i}}{2}$ and \[{{w}^{2}}\ =\ \dfrac{-1-\sqrt{3i}}{2}\] . And it is given that $\alpha $ and $\beta $ are real and $\left( \alpha <\beta \right)$. So, we don’t take values of $\alpha $ and $\beta $ as 1. So, please take care of all the information in the question.