मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता १२

Show that the equation x9 – 5x5 + 4x4 + 2x2 + 1 = 0 has atleast 6 imaginary solutions - Mathematics

Advertisements
Advertisements

प्रश्न

Show that the equation x9 – 5x5 + 4x4 + 2x2 + 1 = 0 has atleast 6 imaginary solutions

बेरीज

उत्तर

P(x) = x9 – 5x5 + 4x4 + 2x2 + 1

(i) The number of sign changes in P(x) is 2.

The number of positive roots is atmost 2.

(ii) P(– x) = – x9 + 5x5 + 4x4 + 2x2 + 1.

The number of sign changes in P(– x) is 1.

The number of negative roots of P(x) is at most 1.

Since the difference of number of sign changes in P(– x) and number of negative zeros is even.

P(x) has one negative root.

(iii) 0 is not the zero of the polynomial P(x).

So the number of real roots is almost 3.

∴ The number of imaginary roots at least 6.

shaalaa.com
Descartes Rule
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Theory of Equations - Exercise 3.6 [पृष्ठ १२७]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 3 Theory of Equations
Exercise 3.6 | Q 3 | पृष्ठ १२७
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×