Question

If the sum of the roots of the quadratic equation $ a{x^2} + bx + c = 0 $ is equal to the sum of the squares of their reciprocals, then prove that $ 2{a^2}c = {c^2}b + {b^2}a $ .

Answer 1

Hint: This is a problem from a quadratic equation. To solve the above question first we will obtain the roots of the quadratic equation and their sum and product respectively. Then using the given data we will derive an equation and by solving that we will get the answer.

Complete step-by-step answer:
The given quadratic equation is, $ a{x^2} + bx + c = 0 $
Let the roots of the given quadratic equation be x and y.
According to the question, the sum of the roots of the quadratic equation $ a{x^2} + bx + c = 0 $ is equal to the sum of the squares of their reciprocals.
 $ \therefore $ $ x + y = \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} $ ………………. (1)
According to the property of quadratic equation, the sum of the roots of the quadratic equation is given by $ - \dfrac{b}{a} $ and product of the roots of the quadratic equation is given by $ \dfrac{c}{a} $ .
Where a is the coefficient of $ {x^2} $ , b is the coefficient of x and c is the constant term in the quadratic equation.
 $ \therefore $ $ x + y = - \dfrac{b}{a} $ and $ xy = \dfrac{c}{a} $
[Though in the question it is asked for sum of the root, but we derived product of the roots also because we will use it later in the solution.]
From equation (1) we have,
 $ x + y = \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} $
Taking L.C.M and simplifying the right hand side of the equation we get,
 $ x + y = \dfrac{{{y^2} + {x^2}}}{{{x^2}{y^2}}} $
Adding and subtracting 2xy in the numerator of the right hand side of the equation we get,
 $ x + y = \dfrac{{{y^2} + {x^2} + 2xy - 2xy}}{{{x^2}{y^2}}} $
As we know the formula, $ {a^2} + {b^2} + 2ab = {(a + b)^2} $ , using it in the above equation we get,
 $ x + y = \dfrac{{{{(y + x)}^2} - 2xy}}{{{x^2}{y^2}}} $
We can write the above equation as,
 $ x + y = \dfrac{{{{(x + y)}^2} - 2xy}}{{{{(xy)}^2}}} $
Above we have derived the values that, $ x + y = - \dfrac{b}{a} $ and $ xy = \dfrac{c}{a} $
Putting these in the above equation we get,
 $ - \dfrac{b}{a} = \dfrac{{{{\left( { - \dfrac{b}{a}} \right)}^2} - 2\dfrac{c}{a}}}{{{{\left( {\dfrac{c}{a}} \right)}^2}}} $
Simplifying the above equation we get,
 $ - \dfrac{b}{a} = \dfrac{{\dfrac{{{b^2}}}{{{a^2}}} - 2\dfrac{c}{a}}}{{\dfrac{{{c^2}}}{{{a^2}}}}} $
Taking denominator of the right hand side to the left hand side we get,
 $ - \dfrac{b}{a} \times \dfrac{{{c^2}}}{{{a^2}}} = \dfrac{{{b^2}}}{{{a^2}}} - 2\dfrac{c}{a} $
Taking L.C.M in right hand side and simplifying the equation we get,
 $ - \dfrac{{b{c^2}}}{{{a^3}}} = \dfrac{{a{b^2} - 2{a^2}c}}{{{a^3}}} $
Cancelling $ {a^3} $ from the denominators of the both side we get,
 $ - b{c^2} = a{b^2} - 2{a^2}c $
By changing sides of the terms we can write the above equation as,
 $ 2{a^2}c = {c^2}b + {b^2}a $
Hence it is proved that if the sum of the roots of the quadratic equation $ a{x^2} + bx + c = 0 $ is equal to the sum of the squares of their reciprocals, then $ 2{a^2}c = {c^2}b + {b^2}a $ .

Note: A quadratic equation is an equation of the second degree i.e. it contains at least one term which is square.
The standard form of quadratic equation is, $ a{x^2} + bx + c = 0 $ , where a, b and c are known values, a can’t be 0. And x is the unknown variable.
To solve the quadratic equation we do the middle term factorization method, graph method, complete the square method and use quadratic formula.
The quadratic formula is given by,
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
And the $ \pm $ means there are two roots.
 $ {b^2} - 4ac = D $ , where D is called discriminant.
When D is positive, we get two real solutions.
When D is zero, we get one real solution (both the answers are the same).
When D is negative, we get a pair of complex solutions.
The sum of all the roots of the quadratic equation is $ \dfrac{{ - b}}{a} $ .
The product of roots of a quadratic equation is $ \dfrac{c}{a} $ .
You should remember all the formulae, rules and properties of quadratic equations.