Answer
Verified
454.8k+ views
Hint: We start solving the by assuming the quadratic equation as ${{x}^{2}}+ax+b=0$. We use the ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\times \left( {{a}^{2}}+{{b}^{2}}+ab \right)$ for difference in cubes of roots and do subsequent calculations to find the value of sum of roots and product of roots. Using this sum and product, we find the quadratic equation.
Complete step-by-step solution:
Given that we have the difference of the roots and difference between cubes of roots of the quadratic equation is 3 and 189. We need to check whether the quadratic equation ${{x}^{2}}\pm 9x+18=0$ or not.
Let us assume the quadratic equation be ${{x}^{2}}+ax+b=0$, and the roots of this quadratic equation be $\alpha $ and $\beta $. We know that sum of the roots $\alpha +\beta =-a$ and product of the roots $\alpha \beta =b$.
According to the problem, we have $\alpha -\beta =3$, and ${{\alpha }^{3}}-{{\beta }^{3}}=189$ -------(1).
We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\times \left( {{a}^{2}}+{{b}^{2}}+ab \right)$. We use this result for equation (1).
We have got ${{\alpha }^{3}}-{{\beta }^{3}}=189$.
We have got $\left( \alpha -\beta \right)\times \left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=189$.
We substitute the value $\alpha -\beta =3$ now.
We have got $3\times \left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=189$.
We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=\dfrac{189}{3}$.
We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=63$ --------(2).
We have got ${{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +\alpha \beta +2\alpha \beta =63$ -------(3).
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. We use this in equation in (3).
So, we have got ${{\left( \alpha -\beta \right)}^{2}}+3\alpha \beta =63$.
We have got ${{3}^{2}}+3\alpha \beta =63$.
We have got $9+3\alpha \beta =63$.
We have got $3\alpha \beta =63-9$.
We have got $3\alpha \beta =54$.
We have got $\alpha \beta =\dfrac{54}{3}$.
We have got $\alpha \beta =18$ -------(4).
From equation (2), we get
We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=63$.
We have got ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta +\alpha \beta -\alpha \beta =63$.
We have got ${{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -\alpha \beta =63$ -------(5).
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We use this in equation in (5).
We have got ${{\left( \alpha +\beta \right)}^{2}}-\alpha \beta =63$.
From equation (4), we use $\alpha \beta =18$.
We have got ${{\left( \alpha +\beta \right)}^{2}}-18=63$.
We have got ${{\left( \alpha +\beta \right)}^{2}}=63+18$.
We have got ${{\left( \alpha +\beta \right)}^{2}}=81$.
We have got $\alpha +\beta =\sqrt{81}$.
We have got $\alpha +\beta =\pm 9$ --------(6).
From equations (5) and (6), we have got $-a=\alpha +\beta =\pm 9$ and $b=\alpha \beta =18$.
Since $\alpha +\beta =\pm 9$, we can take the value of as $\pm 9$.
So, the quadratic equation ${{x}^{2}}\pm 9x+18=0$.
$\therefore$ The required quadratic equation is ${{x}^{2}}\pm 9x+18=0$.
The correct option for the given problem is (a) True.
Note: Here we can take the equation of the quadratic equation is $a{{x}^{2}}+bx+c=0$ and solve for the values of a, b and c by using the sum and product of the roots of the equation. But we need to solve for the value of ‘a’ again. Whenever we get to solve this type of problem, we start by assuming the appropriate quadratic equation.
Complete step-by-step solution:
Given that we have the difference of the roots and difference between cubes of roots of the quadratic equation is 3 and 189. We need to check whether the quadratic equation ${{x}^{2}}\pm 9x+18=0$ or not.
Let us assume the quadratic equation be ${{x}^{2}}+ax+b=0$, and the roots of this quadratic equation be $\alpha $ and $\beta $. We know that sum of the roots $\alpha +\beta =-a$ and product of the roots $\alpha \beta =b$.
According to the problem, we have $\alpha -\beta =3$, and ${{\alpha }^{3}}-{{\beta }^{3}}=189$ -------(1).
We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\times \left( {{a}^{2}}+{{b}^{2}}+ab \right)$. We use this result for equation (1).
We have got ${{\alpha }^{3}}-{{\beta }^{3}}=189$.
We have got $\left( \alpha -\beta \right)\times \left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=189$.
We substitute the value $\alpha -\beta =3$ now.
We have got $3\times \left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=189$.
We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=\dfrac{189}{3}$.
We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=63$ --------(2).
We have got ${{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +\alpha \beta +2\alpha \beta =63$ -------(3).
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. We use this in equation in (3).
So, we have got ${{\left( \alpha -\beta \right)}^{2}}+3\alpha \beta =63$.
We have got ${{3}^{2}}+3\alpha \beta =63$.
We have got $9+3\alpha \beta =63$.
We have got $3\alpha \beta =63-9$.
We have got $3\alpha \beta =54$.
We have got $\alpha \beta =\dfrac{54}{3}$.
We have got $\alpha \beta =18$ -------(4).
From equation (2), we get
We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=63$.
We have got ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta +\alpha \beta -\alpha \beta =63$.
We have got ${{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -\alpha \beta =63$ -------(5).
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We use this in equation in (5).
We have got ${{\left( \alpha +\beta \right)}^{2}}-\alpha \beta =63$.
From equation (4), we use $\alpha \beta =18$.
We have got ${{\left( \alpha +\beta \right)}^{2}}-18=63$.
We have got ${{\left( \alpha +\beta \right)}^{2}}=63+18$.
We have got ${{\left( \alpha +\beta \right)}^{2}}=81$.
We have got $\alpha +\beta =\sqrt{81}$.
We have got $\alpha +\beta =\pm 9$ --------(6).
From equations (5) and (6), we have got $-a=\alpha +\beta =\pm 9$ and $b=\alpha \beta =18$.
Since $\alpha +\beta =\pm 9$, we can take the value of as $\pm 9$.
So, the quadratic equation ${{x}^{2}}\pm 9x+18=0$.
$\therefore$ The required quadratic equation is ${{x}^{2}}\pm 9x+18=0$.
The correct option for the given problem is (a) True.
Note: Here we can take the equation of the quadratic equation is $a{{x}^{2}}+bx+c=0$ and solve for the values of a, b and c by using the sum and product of the roots of the equation. But we need to solve for the value of ‘a’ again. Whenever we get to solve this type of problem, we start by assuming the appropriate quadratic equation.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Who was the Governor general of India at the time of class 11 social science CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference Between Plant Cell and Animal Cell