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Question: If α and β are the roots of the equation\(a{{x}^{2}}+bx+c\), then find the values of (a). \(\dfrac...

If α and β are the roots of the equationax2+bx+ca{{x}^{2}}+bx+c, then find the values of
(a). 1α2+1β2\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}
(b). α4β7+α7β4{{\alpha }^{4}}{{\beta }^{7}}+{{\alpha }^{7}}{{\beta }^{4}}
(c). (αβ−βα)2{{\left( \dfrac{\alpha }{\beta }-\dfrac{\beta }{\alpha } \right)}^{2}}

Explanation

Solution

Hint: For a given quadratic equation of form ax2+bx+ca{{x}^{2}}+bx+cwhose roots are α and β,
the sum of the roots = (α+β)=−ba\left( \alpha +\beta \right)=\dfrac{-b}{a}and the product of the roots =αβ=ca\alpha \beta =\dfrac{c}{a}. We have to substitute these equations in the solution to get our desired answer.

Complete step-by-step answer:

So here we are an equation ax2+bx+ca{{x}^{2}}+bx+cwhose roots are α and β.
We have to find
1α2+1β2\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}
So taking LCM of the equation we get;
α2+β2α2β2\dfrac{{{\alpha }^{2}}\text{+}{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}
=(α+β)2−2.α.β(αβ)2\dfrac{{{(\alpha \text{+}\beta )}^{2}}-2.\alpha .\beta }{{{\left( \alpha \beta \right)}^{2}}} (Becausea2+b2=(a+b)2−2.a.b{{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2.a.b)
= (−ba)2−2.(ca)(ca)2\dfrac{{{\left( \dfrac{-b}{a} \right)}^{2}}-2.\left( \dfrac{c}{a} \right)}{{{\left( \dfrac{c}{a} \right)}^{2}}} (Because α + β = −ba\dfrac{-b}{a} and αβ =ca\dfrac{c}{a})
=b2a2−2.cac2a2\dfrac{\dfrac{{{b}^{2}}}{{{a}^{2}}}-2.\dfrac{c}{a}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}} =b2a2−2cac2a2\dfrac{\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}
Taking a2{{a}^{2}}as LCM in the numerator, we get;
= b2−2aca2c2a2\dfrac{\dfrac{{{b}^{2}}-2ac}{{{a}^{2}}}}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}
= b2−2aca2∗a2c2\dfrac{{{b}^{2}}-2ac}{{{a}^{2}}}*\dfrac{{{a}^{2}}}{{{c}^{2}}} (By reciprocal method)
Cancelling common terma2{{a}^{2}}, we get;
= b2−2acc2\dfrac{{{b}^{2}}-2ac}{{{c}^{2}}}
Therefore the required solution is b2−2acc2\dfrac{{{b}^{2}}-2ac}{{{c}^{2}}}
We have to find
α4β7+α7β4{{\alpha }^{4}}{{\beta }^{7}}+{{\alpha }^{7}}{{\beta }^{4}}
Taking α4β4{{\alpha }^{4}}{{\beta }^{4}}as common, we get;
α4β4(β3+α3){{\alpha }^{4}}{{\beta }^{4}}({{\beta }^{3}}+{{\alpha }^{3}})
= (αβ)4(α+β)(α2+β2−αβ){{\left( \alpha \beta \right)}^{4}}\left( \alpha +\beta \right)\left( {{\alpha }^{2}}+{{\beta }^{2}}-\alpha \beta \right) (Becausea3+b3=(a+b)(a2−ab+b2){{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}}))
= (αβ)4(α+β)((α+β)2−2αβ−αβ){{\left( \alpha \beta \right)}^{4}}\left( \alpha +\beta \right)\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta -\alpha \beta \right) (Becausea2+b2=(a+b)2−2.a.b{{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2.a.b)
= (αβ)4(α+β)((α+β)2−3αβ){{\left( \alpha \beta \right)}^{4}}\left( \alpha +\beta \right)\left( {{\left( \alpha +\beta \right)}^{2}}-3\alpha \beta \right)
We know α + β=−ba\alpha \text{ }+\text{ }\beta =\dfrac{-b}{a} and αβ=ca\alpha \beta =\dfrac{c}{a}
Applying them in the equation, we get;
=(ca)4(−ba)((−ba)2−3ca){{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( {{\left( \dfrac{-b}{a} \right)}^{2}}-\dfrac{3c}{a} \right)
= (ca)4(−ba)((−ba)2−3ca){{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( {{\left( \dfrac{-b}{a} \right)}^{2}}-\dfrac{3c}{a} \right)
=(ca)4(−ba)(b2a2−3ca){{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( \dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{3c}{a} \right)
Taking a2{{a}^{2}} as LCM, we get;
= (ca)4(−ba)(b2−3aca2){{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-b}{a} \right)\left( \dfrac{{{b}^{2}}-3ac}{{{a}^{2}}} \right)
=(ca)4(−b3+3abca3){{\left( \dfrac{c}{a} \right)}^{4}}\left( \dfrac{-{{b}^{3}}+3abc}{{{a}^{3}}} \right)
=(−b3c4+3abc5a7)\left( \dfrac{-{{b}^{3}}{{c}^{4}}+3ab{{c}^{5}}}{{{a}^{7}}} \right)
Therefore the required solution is (−b3c4+3abc5a7)\left( \dfrac{-{{b}^{3}}{{c}^{4}}+3ab{{c}^{5}}}{{{a}^{7}}} \right).
We have to find
(αβ−βα)2{{\left( \dfrac{\alpha }{\beta }-\dfrac{\beta }{\alpha } \right)}^{2}}
Taking LCM we get;
(α2−β2αβ)2{{\left( \dfrac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right)}^{2}}
We knowa2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=(a+b)(a-b). Applying this in the equation we get,
=((α−β)(α+β)αβ)2{{\left( \dfrac{(\alpha -\beta )(\alpha +\beta )}{\alpha \beta } \right)}^{2}}
= (α - β)2(α + β)2(αβ)2\dfrac{{{(\alpha \text{ - }\beta )}^{2}}{{(\alpha \text{ + }\beta )}^{2}}}{{{(\alpha \beta )}^{2}}} …… (i)
We know α + β = −ba\dfrac{-b}{a} and αβ = ca\dfrac{c}{a}
And (α - β)2=(α + β)2−4αβ{{(\alpha \text{ - }\beta )}^{2}}={{(\alpha \text{ + }\beta )}^{2}}-4\alpha \beta
So , (α - β)2=(−ba)2−4(ca){{(\alpha \text{ - }\beta )}^{2}}={{\left( \dfrac{-b}{a} \right)}^{2}}-4\left( \dfrac{c}{a} \right)
= b2a2−4ca\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{4c}{a}
Taking a2{{a}^{2}}as LCM, we get;
(α - β)2{{(\alpha \text{ - }\beta )}^{2}} =b2−4aca2\dfrac{{{b}^{2}}-4ac}{{{a}^{2}}}
Putting the value of (α - β)2{{(\alpha \text{ - }\beta )}^{2}} in equation (i), we get;
(b2−4ac)a2.(−ba)(ca)22{{\dfrac{\dfrac{\left( {{b}^{2}}-4ac \right)}{{{a}^{2}}}.\left( \dfrac{-b}{a} \right)}{{{\left( \dfrac{c}{a} \right)}^{2}}}}^{2}}
= (b2−4ac)a2.(b2a2)c2a2\dfrac{\dfrac{\left( {{b}^{2}}-4ac \right)}{{{a}^{2}}}.\left( \dfrac{{{b}^{2}}}{{{a}^{2}}} \right)}{\dfrac{{{c}^{2}}}{{{a}^{2}}}}
Multiplying and taking reciprocal we get;
b4−4acb2a4.a2c2\dfrac{{{b}^{4}}-4ac{{b}^{2}}}{{{a}^{4}}}.\dfrac{{{a}^{2}}}{{{c}^{2}}}
Cancelling the common term a2{{a}^{2}}we get;
b4−4acb2a2c2\dfrac{{{b}^{4}}-4ac{{b}^{2}}}{{{a}^{2}}{{c}^{2}}}
So, the required answer isb4−4acb2a2c2\dfrac{{{b}^{4}}-4ac{{b}^{2}}}{{{a}^{2}}{{c}^{2}}}.

Note: It is very important to remember the formulas of
α+β=−ba\alpha \text{+}\beta =\dfrac{-b}{a} , αβ=ca\alpha \beta =\dfrac{c}{a} and (α−β)2=b2−4aca2{{\left( \alpha -\beta \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{{{a}^{2}}} many times they will be used in quadratic equation problems. We can also write α−β=∣D∣a\alpha -\beta =\dfrac{|\sqrt{D}|}{a} where D is the discriminant where Discriminant D=b2−4acD={{b}^{2}}-4ac.