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Question: If \[\alpha \] and \[\beta \] are the zeros of the quadratic polynomial \[p(s)=3{{s}^{2}}-6s+4\], fi...

If α\alpha and β\beta are the zeros of the quadratic polynomial p(s)=3s2−6s+4p(s)=3{{s}^{2}}-6s+4, find the value of αβ+βα+2(1α+1β)+3αβ\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta .

Explanation

Solution

Hint: We will use the formula of summation of two roots that is α+β=−ba\alpha +\beta =\dfrac{-b}{a} and product of two roots that is αβ=ca\alpha \beta =\dfrac{c}{a} to solve this question. Then using these two we find α2+β2{{\alpha }^{2}}+{{\beta }^{2}} and then substitute the values of all these terms in the given expression.

Complete step-by-step solution -
Quadratic polynomial given in the question is p(s)=3s2−6s+4......(1)p(s)=3{{s}^{2}}-6s+4......(1) has two roots(zeros), namely α\alpha and β\beta from coefficient and zeros relation.
We know that the summation of the two zeros is α+β=−ba........(2)\alpha +\beta =\dfrac{-b}{a}........(2)
Also the product of the two zeros is αβ=ca........(3)\alpha \beta =\dfrac{c}{a}........(3)
Now comparing equation (1) with the general form of quadratic polynomial p(s)=as2+bs+cp(s)=a{{s}^{2}}+bs+c we get a as 3, b as -6 and c as 4. Now substituting these values of coefficients in equation (2) and equation (3) we get,
α+β=−(−6)3=63=2........(4)\alpha +\beta =\dfrac{-(-6)}{3}=\dfrac{6}{3}=2........(4) and αβ=43........(5)\alpha \beta =\dfrac{4}{3}........(5)
Now we will solve this expression αβ+βα+2(1α+1β)+3αβ........(6)\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta ........(6)
Rearranging and simplifying equation (6) we get,
 ⇒α2+β2αβ+2(α+βαβ)+3αβ........(7)\,\Rightarrow \dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }+2\left( \dfrac{\alpha +\beta }{\alpha \beta } \right)+3\alpha \beta ........(7)
In equation (7) the value of the term α2+β2{{\alpha }^{2}}+{{\beta }^{2}} is not known and all other terms value is known. So now finding the value of α2+β2{{\alpha }^{2}}+{{\beta }^{2}}.
From equation (4) we get,
⇒α+β=2\Rightarrow \alpha +\beta =2
Now squaring both sides of this equation we get,
 ⇒α2+β2+2αβ=4......(8)\,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta =4......(8)
Substituting the value of αβ\alpha \beta from equation (5) in equation (8) we get,

& \,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+2\times \dfrac{4}{3}=4 \\\ & \,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=4-\dfrac{8}{3} \\\ & \,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=\dfrac{4}{3}........(9) \\\ \end{aligned}$$ Now substituting all the values from equation (4), equation (5) and equation (9) in equation (7) we get, $$\begin{aligned} & \,\Rightarrow \dfrac{\dfrac{4}{3}}{\dfrac{4}{3}}+2\left( \dfrac{2}{\dfrac{4}{3}} \right)+3\times \dfrac{4}{3} \\\ & \,\Rightarrow 1+3+4=8 \\\ \end{aligned}$$ Hence the answer to this expression $$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta $$ is 8. Note: We have to remember both the formulas of sum of roots and product of the roots because these are the key to solve this problem. We may get confused in equation (7) about how to find $${{\alpha }^{2}}+{{\beta }^{2}}$$ but here we have can find it by squaring both sides of $$\Rightarrow \alpha +\beta =2$$. In a hurry we may make simple calculation mistakes, so we need to be very careful with every step.