Question

If \[\alpha \]and \[\beta \] are the roots of the equation \[{x^2} - 7x + 1 = 0\] then the value of \[\dfrac{1}{{{{\left( {\alpha - 7} \right)}^2}}} + \dfrac{1}{{{{\left( {\beta - 7} \right)}^2}}}\] is
A. \[45\]
B. \[47\]
C. \[49\]
D. \[50\]
E. \[51\]

Answer 1

Hint: We know that any quadratic equation in the variable \[x\] is of the form \[a{x^2} + bx + c = 0\] . Comparing the standard equation with the given equation, we can find the variables and then we can use the formula to find the sum of the roots \[ = - \dfrac{b}{a}\] and the product of roots \[ = \dfrac{c}{a}\] .

Complete step by step answer:
A polynomial is an expression that can be built from constants and symbols called variables or indeterminants by means of addition, multiplication and exponentiation to a non-negative integer power
A polynomial in a single indeterminate x can always be written (or rewritten) in the form
\[{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + .......{a_2}{x^2} + {a_1}x + {a_0}\] where \[{a_0},{a_1}......,{a_n}\] are constants and \[x\] is the indeterminate. The word “indeterminate” means that \[x\] represents no particular value , although any value may be substituted for it.
We know that any quadratic equation in the variable \[x\] is of the form \[a{x^2} + bx + c = 0\] .
We are given the equation \[{x^2} - 7x + 1 = 0\]
On comparing both the equations we get \[a = 1,b = - 7,c = 1\]
Using the properties of roots that are as follows
Sum of the roots \[ = - \dfrac{b}{a}\]
Product of the roots \[ = \dfrac{c}{a}\]
Therefore we get
\[\alpha \beta = \dfrac{1}{1} = 1\]
Squaring both the sides we get
             \[{\alpha ^2}{\beta ^2} = 1\]
Sum of the roots of quadratic equation
\[\alpha + \beta = - \dfrac{{ - 7}}{1} = 7\]
Using the property \[{\left( {\alpha + \beta } \right)^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta \]
Therefore we get ,
\[{\left( 7 \right)^2} = {\alpha ^2} + {\beta ^2} + 2\]
On simplifying we get ,
\[47 = {\alpha ^2} + {\beta ^2}\]
Now consider ,
\[\dfrac{1}{{{{\left( {\alpha - 7} \right)}^2}}} + \dfrac{1}{{{{\left( {\beta - 7} \right)}^2}}}\]
On taking the LCM we get ,
\[ = \dfrac{{{{\left( {\beta - 7} \right)}^2} + {{\left( {\alpha - 7} \right)}^2}}}{{{{\left( {\beta - 7} \right)}^2}{{\left( {\alpha - 7} \right)}^2}}}\]
On solving this we get ,
\[ = \dfrac{{{\beta ^2} - 14\beta + 49 + {\alpha ^2} - 14\alpha + 49}}{{\left( {{\beta ^2} - 14\beta + 49} \right)\left( {{\alpha ^2} - 14\alpha + 49} \right)}}\]
On simplifying the denominator we get ,
\[ = \dfrac{{{\alpha ^2} + {\beta ^2} - 14\left( {\alpha + \beta } \right) + 98}}{{\left( {{\alpha ^2}{\beta ^2} - 14\alpha {\beta ^2} + 49{\beta ^2} - 14{\alpha ^2}\beta + 196\alpha \beta - 686\beta + 49{\alpha ^2} - 686\alpha + 2401} \right)}}\]
Which on further simplification becomes
\[ = \dfrac{{47 - 14\left( 7 \right) + 98}}{{\left( {1 - 14\alpha \beta (\alpha + \beta ) + 49({\alpha ^2} + {\beta ^2}) + 196 - 686(\alpha + \beta ) + 2401} \right)}}\]
Which on further simplification becomes
\[ = \dfrac{{47 - 14\left( 7 \right) + 98}}{{\left( {1 - 14(7) + 49(47) + 196 - 686(7) + 2401} \right)}}\]
Which on further simplification becomes
\[ = \dfrac{{47 - 98 + 98}}{{\left( {1 - 98 + 2303 + 196 - 4802 + 2401} \right)}}\]
Which finally gives us the answer
\[ = 47\]

So, the correct answer is “Option B”.

Note: We can also find the solution by finding the roots individually using the Shridhar Acharya’s formula and substituting in the given equation. The formulas we have used here will only be applicable to the quadratic equations.