Solveeit Logo
Login

Question

Question: If \(\alpha \) and \(\beta \) are the roots of \(a{x^2} + bx + c = 0\), find the values of the follo...

If α\alpha and β\beta are the roots of ax2+bx+c=0a{x^2} + bx + c = 0, find the values of the following. βaα+b+αaβ+b\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}}

Explanation

Solution

Hint: Before approaching this question prior knowledge of quadratic equation is must, put the values of α\alpha and β\beta in the quadratic equation and make the equation, use this information to approach towards the solution to the problem

Complete step-by-step solution -
According to the given information we have quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 which have roots α\alpha and β\beta
Substituting the value of x = α\alpha in the given quadratic equation i.e. ax2+bx+c=0a{x^2} + bx + c = 0 we get
a(α)2+b(α)+c=0a{\left( \alpha \right)^2} + b\left( \alpha \right) + c = 0
\Rightarrow $$$a{\alpha ^2} + b\alpha = - c$$ …………….(eq.1) Now for x =\beta $$a{\left( \beta \right)^2} + b\left( \beta \right) + c = 0$$ \Rightarrow $aβ2+bβ=−ca{\beta ^2} + b\beta = - c …………..(eq.2)
As we know that both roots of a quadratic equation so for the roots of quadratic equation α+β=−ba\alpha + \beta = \dfrac{{ - b}}{a} and αβ=ca\alpha \beta = \dfrac{c}{a}
Now finding the value of βaα+b+αaβ+b\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}}
Now substituting the given values in the above equation
βaα+b+αaβ+b=β(aβ+b)+α(aα+b)(aβ+b)(aα+b)\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{\beta \left( {a\beta + b} \right) + \alpha \left( {a\alpha + b} \right)}}{{\left( {a\beta + b} \right)\left( {a\alpha + b} \right)}}
⇒\Rightarrow βaα+b+αaβ+b=aβ2+bβ+aα2+bαa2αβ+abα+abβ+b2\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{a{\beta ^2} + b\beta + a{\alpha ^2} + b\alpha }}{{{a^2}\alpha \beta + ab\alpha + ab\beta + {b^2}}} …………(eq. 3)
Substituting the given values and values form equation 1 and equation 2 to equation 3
⇒\Rightarrow βaα+b+αaβ+b=−c−ca2(ca)+ab(−ba)+b2\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - c - c}}{{{a^2}\left( {\dfrac{c}{a}} \right) + ab\left( {\dfrac{{ - b}}{a}} \right) + {b^2}}}
⇒\Rightarrow βaα+b+αaβ+b=−2cac−b2+b2\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - 2c}}{{ac - {b^2} + {b^2}}}
⇒\Rightarrow βaα+b+αaβ+b=−2cac\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - 2c}}{{ac}}
⇒\Rightarrow βaα+b+αaβ+b=−2a\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - 2}}{a}
Hence the value of βaα+b+αaβ+b\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} is −2a\dfrac{{ - 2}}{a}.

Note: In the above question which was based on quadratic equation so let’s discuss about the quadratic equation which can be explained as the equation which consist of at least one squared term also this equation is called as second degree equation. The general representation of quadratic equation is ax2+bx+c=0a{x^2} + bx + c = 0 where a, b and c are the constant of quadratic equation and x represents the unknown variable.