Question

If ‘r’ and ‘s’ are the roots of the equation \[a{{x}^{2}}+bx+c=0\], then the value of \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\] is equal to
a) \[{{b}^{2}}-4ac\]
b) \[\dfrac{{{b}^{2}}-4ac}{2a}\]
c) \[\dfrac{{{b}^{2}}-4ac}{{{c}^{2}}}\]
d) \[\dfrac{{{b}^{2}}-2ac}{{{c}^{2}}}\]

Answer 1

Hint: In this question, we can approach by finding out the sum of the roots, that is, \[r+s\] and product of the roots, that is, \[r\times s\], and then we can convert \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\] in terms of sum and product of ‘r’ and ‘s’.

Complete step by step answer:
In this question, we are given with a quadratic equation \[a{{x}^{2}}+bx+c=0\], and it is given that the roots of the given quadratic equation are ‘r’ and ‘s’.
We have to find the value of \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\], to find this, we will simplify this in such a manner, so that it can be represented in terms of sum and product of ‘r’ and ‘s’, that is, \[r+s\] and \[r\times s\] respectively.
As we know that , sum of roots of a quadratic equation like \[a{{x}^{2}}+bx+c=0\] can be written as equal to \[-\dfrac{b}{a}\] and product of roots of the equation is always equal to \[\dfrac{c}{a}\].
This means, sum of roots, \[r+s=-\dfrac{b}{a}......\left( i \right)\], and product of roots, \[r\times s=\dfrac{c}{a}......\left( ii \right)\].
Now, we can simplify \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\] by simply taking L.C.M.,
That is, \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\] can be written as
\[\dfrac{{{r}^{2}}+{{s}^{2}}}{{{r}^{2}}\times {{s}^{2}}}\]
\[\Rightarrow \dfrac{{{r}^{2}}+{{s}^{2}}}{{{\left( r\times s \right)}^{2}}}......\left( iii \right)\]
As we know that, \[{{\left( r+s \right)}^{2}}={{r}^{2}}+{{s}^{2}}+2r\times s\]
Therefore, we can write \[{{r}^{2}}+{{s}^{2}}={{\left( r+s \right)}^{2}}-2r\times s......\left( iv \right)\]
Now, from using equation (iii) and (iv), we get,
\[\dfrac{{{r}^{2}}+{{s}^{2}}}{{{\left( r\times s \right)}^{2}}}=\dfrac{{{\left( r+s \right)}^{2}}-2r\times s}{{{\left( r\times s \right)}^{2}}}\]
\[\Rightarrow \dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}=\dfrac{{{\left( r+s \right)}^{2}}-2r\times s}{{{\left( r\times s \right)}^{2}}}......\left( v \right)\]
Here, we can see that we have represent \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\] in terms of sum and product of roots, that is, \[r+s\] and \[r\times s\] respectively.
Now, to get the value of \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\], we will put the value of \[r+s\] and \[r\times s\] from equation (i) and (ii), to (v).
Therefore, we get,
\[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}=\dfrac{{{\left( -\dfrac{b}{a} \right)}^{2}}-2\dfrac{c}{a}}{{{\left( \dfrac{c}{a} \right)}^{2}}}\]
Now, we will simplify both numerator and denominator
\[\Rightarrow \dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}=\dfrac{\dfrac{{{\left( b \right)}^{2}}-2ac}{{{a}^{2}}}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}\]
\[\Rightarrow \dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}=\dfrac{\left( {{\left( b \right)}^{2}}-2ac \right){{a}^{2}}}{{{c}^{2}}{{a}^{2}}}\]
\[\Rightarrow \dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}=\dfrac{\left( {{\left( b \right)}^{2}}-2ac \right)}{{{c}^{2}}}\]
Therefore, option (d) is correct.

Note: We can also use another way to find the value of \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}\] without taking L.C.M. by assuming \[\dfrac{1}{r}\] as one term and \[\dfrac{1}{s}\] as second term and then, \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}={{\left( \dfrac{1}{r}+\dfrac{1}{s} \right)}^{2}}-\dfrac{2}{rs}\], from equation (iv), and then by taking L.C.M., we will get, \[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{s}^{2}}}={{\left( \dfrac{r+s}{r\times s} \right)}^{2}}-\dfrac{2}{rs}\], and then we can simply put the values of \[r+s\] and \[r\times s\] from equation (i) and (ii) and then we can simplify it.