Question

How many real solutions does the equation

x⁷ + 14x⁵ + 16x³ + 30x –560 = 0 have?

• A. 1
• B. 3
• C. 5
• D. 7

Answer 1

Answer: (A)

1

Answer 2

Here f(x) = x⁷ + 14x⁵ + 16x³ + 30x – 560

Since degree is odd so at least one solution is real.

f ¢(x) = 7x⁶ + 70x⁴ + 48x² + 30 ³ 30 " x Ī R as all the powers are even.

so f(x) is increasing " x so only one real root.