Question

If a1, a2, a3, .......an (n >= 2) are real and (n-1)a1² - 2na2<0 then prove that at least two
roots of the equation x^n + a2x^(n-2) + an = 0 are imaginary.

Please tell urgent .
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Answer 1

Step-by-step explanation:

let                                                                                                                                          

                     $f(x) = x^k + a_1x^(^k^-^1^) + a_2x^(^k^-^2^) + ... + a_k = 0$

now,

if this function has at least two imaginary roots, Then the derivative of this function when the highest degree becomes two should have 2 non-real roots.

so,the (K-2) th derivative is given by:                                                                                          

                                                                                                                                                                                                                      $f^(^k^-^2^)(x) = (k!)x^2 + ((k-1)!) a_1x + ((k-2)!)a2 =0$

using the rule                       $\frac{d}{dx} (x^n) = nx^n^-^1$

now, after taking ((k-2)!) common and cancelling on both sides, we are left with:                          

                $f^(^k^-^2^)(x) =k(k-1)x^2+a_1(k-1)x+a_2=0$ ,

which is a quadratic equation in x. NOW, for this equation, we get the discriminant, D as:            

                              $D=b^2-4ac$

                                  $= a_1^2 (k-1)^2 - 4(k)(k-1)a_2$

                                  $= (k-1)((k-1)a_1^2 -4ka_2)$

now, we know that since k>2 , k-1 > 0 and it was initially given that

                                   $(k-1)a_1^2 -4ka_2 < 0$

SO, we get D<0 , which means  $f^(^k^-^2^)$ (x) has 2 imaginary roots, which in turn means that f(x) has at least 2 imaginary roots