Solve the given equations.
$4{{x}^{3}}+20{{x}^{2}}-23x+6=0$ , Two of the roots being equal.
VerifiedHint: Assume a cubic with 2 equal roots. Write it as a product of 3 terms of x. Solve that equation to compare the coefficient to get values of roots. Assume variables a, b, c as coefficients. The values of x are required to result in this question.
Complete step-by-step answer:
Given equation in the question, can be written in form of
$4{{x}^{3}}+20{{x}^{2}}-23x+6=0$
Let us take the equal roots to be denoted by a. So, we can write the function in terms of x as
\[(x-a)(x-a)(cx-b)=0\]
\[\left( cx-b \right)\] term indicates the ${{3}^{rd}}$ root, which is not equal. By simplifying the above equation, we can write it as
$({{x}^{2}}-2ax+{{a}^{2}})(cx-b)=0$
By assuming $(cx-b)$ as k and simplify we get it as
$k{{x}^{2}}-2akx+k{{a}^{2}}=0$
By substituting value of k back into equation, we get
$(cx-b){{x}^{2}}-2a(cx-b)x+(cx-b){{a}^{2}}=0$
By simplifying the term, we can write it as
$c{{x}^{3}}-b{{x}^{2}}-2ac{{x}^{2}}+2abx+{{a}^{2}}cx-b{{a}^{2}}=0$
By grouping the terms together, we get it in the form of
$c{{x}^{3}}-2ac{{x}^{2}}-b{{x}^{2}}+2abx+{{a}^{2}}cx-b{{a}^{2}}=0$
By simplifying the above equation, we can write it in the form of
$c{{x}^{3}}-(2ac+b){{x}^{2}}+(2ab+{{a}^{2}}c)x-b{{a}^{2}}=0$
The given equation in question is written here, we get it as
$4{{x}^{3}}+20{{x}^{2}}-23x+6=0$
By comparing, we get value of c directly as given below
\[c=4\] …..( 1 )
By equating the second terms, we can write it as\[-(2ac+b)=20\] . By substituting value of c, we can write it as
\[8a+b=-20\]…….( 2 )
By subtracting 8a on both sides, we get it as
\[b=-20-8a\] ……( 3 )
By comparing the third term, we can write it as
$2ab+{{a}^{2}}c=-23$
By substituting the equation (3), we get it as
$2a(-20-8a)+4{{a}^{2}}=-23$
By simplifying we can say the equations turns into
\[-40a-16{{a}^{2}}+4{{a}^{2}}=-23\]
By simplifying whole equation, we can write it as
$12{{a}^{2}}+40a-23=0$
By writing $-40a$ as $-6a+46a$, we get it as
$12{{a}^{2}}-6a+46a-23=0$
By taking 6a common from first 2 terms, we get it as
\[6a(2a-1)+46a-23=0\]
By taking 23 common from last 2 terms, we get it as
\[6a(2a-1)+23(2a-1)=0\]
By taking \[(2a-1)\] as common, we can write it as
\[(2a-1)(6a+23)=0\]
By solving them, we can say values of a as given below
\[a=\dfrac{1}{2},a=-\dfrac{23}{6}\]
If \[a=\dfrac{1}{2}\] is substituted in (3), we get it in form
$b=-20-8\left( \dfrac{1}{2} \right)=-24$
If \[a=-\dfrac{23}{6}\] is substituted in (3), we get it in form
$b=-20-8\left( -\dfrac{23}{6} \right)=\dfrac{72}{3}$
By comparing constants, we get equation in the form of
${{a}^{2}}b=-6$
This is satisfied by the pair \[\left( \dfrac{1}{2},-24 \right)\] . So, this is taken. By substituting a, b, c we get the equation as
$\left( x-\dfrac{1}{2} \right)\left( x-\dfrac{1}{2} \right)(4x+24)=0$
So, the roots are $\dfrac{1}{2},\dfrac{1}{2},-6$ for a given equation.
Note: Be careful with the values you take a, b, c because more variables will make the problem tougher. So, take the variables as least as possible. Whenever you see a cubic you can’t use this method. This method is advisable only in case of equal roots as there are less variables in this case.
