Answer
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Hint: First we will expand the given equation and we will get the equation to be as $\left( 1+a \right){{x}^{2}}+2x-3=0$ . Then we will assume the roots of this equation to be ${{x}_{1}},{{x}_{2}}$ . Also, we will find the sum and product of roots given as $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ respectively. Then there is a rule, if one root tends to infinity, it is only possible when the coefficient of ${{x}^{2}}$ must tend to zero and coefficient of x must not be equal to 0. Using this, we will find the value of a.
Complete step-by-step answer:
We are given with the equation $x\left( x+2 \right)=3-a{{x}^{2}}$ which is in quadratic form. So, on further simplification we can write it as
${{x}^{2}}+2x=3-a{{x}^{2}}$
On taking all the terms on left hand side, we get as
${{x}^{2}}+a{{x}^{2}}+2x-3=0$
Now taking ${{x}^{2}}$ common, we get as
$\left( 1+a \right){{x}^{2}}+2x-3=0$ …………………(1)
Now, let us assume that the roots of this quadratic equation are ${{x}_{1}},{{x}_{2}}$ . We also know that the sum of the roots are equal to $\dfrac{-b}{a}$ and products of the roots are equal to $\dfrac{c}{a}$ .
So, we will find the same for equation (1). We get as
Sum of the roots ${{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}=\dfrac{-2}{1+a}$ where b is coefficient of x term and a is co-efficient of ${{x}^{2}}$ .
Product of roots ${{x}_{1}}{{x}_{2}}=\dfrac{c}{a}=\dfrac{-3}{1+a}$ where c is the constant term in equation.
Now, we are given that one of the either roots tends to infinity. It is only possible when the coefficient of ${{x}^{2}}$ must tend to zero and the coefficient of x must not be equal to 0. So, we can say that the coefficient of ${{x}^{2}}$ i.e. $1+a=0$ .
On solving this we will get $a=-1$ .
Thus, option (b) is correct answer.
Note: Another method is by doing option method. After getting a quadratic equation, we can directly put values of ‘a’ in the equation and see which values make the coefficient zero, that will be the answer. So, this way also we will get the answer. Also, students should know the concept when roots tend to infinity then what are the criteria which needs to be done then only the answer will be correct. So, be careful with all these details while solving.
Complete step-by-step answer:
We are given with the equation $x\left( x+2 \right)=3-a{{x}^{2}}$ which is in quadratic form. So, on further simplification we can write it as
${{x}^{2}}+2x=3-a{{x}^{2}}$
On taking all the terms on left hand side, we get as
${{x}^{2}}+a{{x}^{2}}+2x-3=0$
Now taking ${{x}^{2}}$ common, we get as
$\left( 1+a \right){{x}^{2}}+2x-3=0$ …………………(1)
Now, let us assume that the roots of this quadratic equation are ${{x}_{1}},{{x}_{2}}$ . We also know that the sum of the roots are equal to $\dfrac{-b}{a}$ and products of the roots are equal to $\dfrac{c}{a}$ .
So, we will find the same for equation (1). We get as
Sum of the roots ${{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}=\dfrac{-2}{1+a}$ where b is coefficient of x term and a is co-efficient of ${{x}^{2}}$ .
Product of roots ${{x}_{1}}{{x}_{2}}=\dfrac{c}{a}=\dfrac{-3}{1+a}$ where c is the constant term in equation.
Now, we are given that one of the either roots tends to infinity. It is only possible when the coefficient of ${{x}^{2}}$ must tend to zero and the coefficient of x must not be equal to 0. So, we can say that the coefficient of ${{x}^{2}}$ i.e. $1+a=0$ .
On solving this we will get $a=-1$ .
Thus, option (b) is correct answer.
Note: Another method is by doing option method. After getting a quadratic equation, we can directly put values of ‘a’ in the equation and see which values make the coefficient zero, that will be the answer. So, this way also we will get the answer. Also, students should know the concept when roots tend to infinity then what are the criteria which needs to be done then only the answer will be correct. So, be careful with all these details while solving.
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