Question

find the zeros of polynomial p(x)=4x^2+11x+10​

Answer 1

We have to find the polynomial of P(x) = 4x² + 11x + 10

Solution : here P(x) = 4x² + 11x + 10

⇒4x² + 11x + 10 = 0

⇒x² + (11/4)x + (10/4) = 0

⇒x² + 2(11/8)x = -(10/4)

⇒x² + 2(11/8)x + (11/8)² = -(10/4) + (11/8)²

⇒(x + 11/8)² = (-160 + 121)/64

⇒(x + 11/8)² = -39/64

Here you see LHS always will be positive while RHS is negative.

So, given polynomial doesn't have real zeroes. You can understand finding D = b² - 4ac

D = (11)² - 4(4)(10) = 121 - 160 = -39 < 0 so can't be real zeroes.

Now if you want to find imaginary zeroes take -1 = i²

So, (x + 11/8)² = 39i²/64

then, x + 11/8 = ±√39i/8

x = -11/8 ± √39i/8

x = (-11 ± √39i)/8

therefore the imaginary zeroes are (-11+√39i)/8 and (-11-√39i)/8