Question

If k√-1 is a root of the equation
x^4 +6x^3-16x^2+24x-80=0 then k=​

Answer 1

Given info : k√(-1) is a root of the equation x⁴ + 6x³ - 16x² + 24x - 80 = 0

To find : the value of k = ?

solution : we know, √(-1) is an imaginary number and it is denoted by iota (i)

so, ki is a root of the equation x⁴ + 6x³ - 16x² + 24x - 80 = 0

so putting the value of x, will satisfy the equation.

(ki)⁴ + 6(ki)³ - 16(ki)² + 24(ki) - 80 = 0

⇒k⁴i⁴ + 6k³i³ - 16k²i² + 24ki - 80 = 0

we know, i² = -1

⇒k⁴(1) + 6k³(-1 × i) + 16k² + 24ki - 80 = 0

⇒k⁴ - 6k³i + 16k² + 24ki - 80 = 0

let's focus on imaginary part of the given equation.

only if this equation will give zero when (-6k³i + 24ki) be zero , right ?

now, -6k³i + 24ki = -6ki(k² - 4) = 0

⇒k = ±2

Therefore the value of k = +2 and -2.

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