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Solving a Quadratic Equation by Factorization Method

It is a KVPY ...

Question

It is a KVPY question Let a1,a2,a3,a4 be real number such that a1+a2+a3+a4=0 and a1^2+a2^2+a3^2+a4^2=1 then the smallest possible value of the expression (a1-a2)^2+(a2-a3)^2+(a3-a4)^2+(a4-a1)^2 lies in which interval. Not sure about the chapter..

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Solution

Since
a1^2 +a2^2+a3^2+a4^2=1

Smallest possible value of

(a1 – a2)^2 + (a2 – a3)^2 + (a3 – a4)^2 + (a4– a1)^2 =0

if a1 = a2 = a3 = a4=( +/-) 1/2

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Similar questions

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

a_{1}, a_{2}, a_{3}, a_{4}

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

(a_{1} - a_{2})^{2} + (a_{2} - a_{2})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

a_{1}, a_{2}, a_{3}, a_{4}

a_{1} + a_{2} + a_{3} + a_{4} = 0

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

(a_{1} - a_{2})^{2} + (a_{2} - a_{3})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

a_{1}, a_{2},

a_{n}

a_{1} = 0

\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \frac{a_{5}}{a_{4}} +

+ \frac{a_{n}}{a_{n - 1}} - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n - 2}}) = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}

a_{1}, a_{2}, a_{3}, a_{4}

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

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