The quadratic equation, whose roots are three times the roots of $$3a{x^2} + 3bx + c = 0$$, is
A.$$a{x^2} + 3bx + 3c = 0$$
B.$$a{x^2} + 3bx + c = 0$$
C.$$9a{x^2} + 9bx + c = 0$$
D.$$a{x^2} + bx + 3c = 0$$
Answer
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Hint: Here in this question, we have to find the quadratic equation by using a given condition. For this, given a condition the unknown quadratic equation roots are three times the roots of $$3a{x^2} + 3bx + c = 0$$ then substitute the $$x = \dfrac{y}{3}$$ in given quadratic equation and on further simplification to get the required solution.
Complete step-by-step answer:
A quadratic equation in x is an equation that can be written in the standard form $$a{x^2} + bx + c = 0$$, where a, b and c are real numbers and $$a \ne 0$$.
‘a’ represents the numerical coefficient of $${x^2}$$,
‘b’ represents the numerical coefficient of $$x$$, and
‘c’ represents the constant numerical term.
Consider the given question,
Find the quadratic equation, whose roots are three times the roots of quadratic equation $$3a{x^2} + 3bx + c = 0$$.
Let us take $$\alpha $$ and $$\beta $$ be the roots of the equation
$$3a{x^2} + 3bx + c = 0$$ ----(1)
We have to find the equation whose roots are $$3\alpha $$ and $$3\beta $$, which can be got by putting $$y = 3x$$ in the given equation, i.e.,
Substituting $$x$$ for $$\dfrac{y}{3}$$ in the given equation (1), then we have
$$ \Rightarrow \,\,\,3a{\left( {\dfrac{y}{3}} \right)^2} + 3b\left( {\dfrac{y}{3}} \right) + c = 0$$
$$ \Rightarrow \,\,\,3a\dfrac{{{y^2}}}{{{3^2}}} + 3b\left( {\dfrac{y}{3}} \right) + c = 0$$
$$ \Rightarrow \,\,\,a\dfrac{{{y^2}}}{3} + by + c = 0$$
Take 3 as LCM in LHS, then we have
$$ \Rightarrow \,\,\,\dfrac{{a{y^2} + 3by + 3c}}{3} = 0$$
Multiply both side by 3, then we have
$$ \Rightarrow \,\,\,a{y^2} + 3by + 3c = 0$$
Replace $$y$$ by $$x$$
$$\therefore \,\,\,\,\,a{x^2} + 3bx + 3c = 0$$
Hence, it’s a required solution.
Therefore, option (1) is the correct answer.
So, the correct answer is “Option 1”.
Note: The general form of quadratic equation is represented as $$a{x^2} + bx + c = 0$$ it represents the polynomial equation whose variable term has degree 2. To simplify we use the simple arithmetic operations like addition, subtraction, multiplication and division and we should know about the tables of multiplication.
Complete step-by-step answer:
A quadratic equation in x is an equation that can be written in the standard form $$a{x^2} + bx + c = 0$$, where a, b and c are real numbers and $$a \ne 0$$.
‘a’ represents the numerical coefficient of $${x^2}$$,
‘b’ represents the numerical coefficient of $$x$$, and
‘c’ represents the constant numerical term.
Consider the given question,
Find the quadratic equation, whose roots are three times the roots of quadratic equation $$3a{x^2} + 3bx + c = 0$$.
Let us take $$\alpha $$ and $$\beta $$ be the roots of the equation
$$3a{x^2} + 3bx + c = 0$$ ----(1)
We have to find the equation whose roots are $$3\alpha $$ and $$3\beta $$, which can be got by putting $$y = 3x$$ in the given equation, i.e.,
Substituting $$x$$ for $$\dfrac{y}{3}$$ in the given equation (1), then we have
$$ \Rightarrow \,\,\,3a{\left( {\dfrac{y}{3}} \right)^2} + 3b\left( {\dfrac{y}{3}} \right) + c = 0$$
$$ \Rightarrow \,\,\,3a\dfrac{{{y^2}}}{{{3^2}}} + 3b\left( {\dfrac{y}{3}} \right) + c = 0$$
$$ \Rightarrow \,\,\,a\dfrac{{{y^2}}}{3} + by + c = 0$$
Take 3 as LCM in LHS, then we have
$$ \Rightarrow \,\,\,\dfrac{{a{y^2} + 3by + 3c}}{3} = 0$$
Multiply both side by 3, then we have
$$ \Rightarrow \,\,\,a{y^2} + 3by + 3c = 0$$
Replace $$y$$ by $$x$$
$$\therefore \,\,\,\,\,a{x^2} + 3bx + 3c = 0$$
Hence, it’s a required solution.
Therefore, option (1) is the correct answer.
So, the correct answer is “Option 1”.
Note: The general form of quadratic equation is represented as $$a{x^2} + bx + c = 0$$ it represents the polynomial equation whose variable term has degree 2. To simplify we use the simple arithmetic operations like addition, subtraction, multiplication and division and we should know about the tables of multiplication.
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