Question

prove that: 4z square -24z +136 = 0

Answer 1

Given: 4z² - 24z + 136 =0

This is simplified as 4(z²-6z+34) = 0

⇒ z²-6z+34 = 0

⇒ Using formula x= [-b±√(b²-4ac)]/(2a) for an equatiion ax²+bx+c=0

here a=1;b=-6,c=34

so z = [-(-6)±√(6²-4*1*34)]/(2*1)

      = [6±√(-100)]/2

      =3±5√(-1)

So z = 3±5*i is a complex number, where i =√(-1) is an imaginary number

Now we substitute z=3±5*i  in the above equation.

So z²= (3±5*i)²= 9-25 ± 2(3)(5*i) = -16±30*i

Now z²-6z+34 = (-16±30*i) - 6(3±5*i ) + 34

                       = (-16-18 + 34) ± i*(30-30)

                       = 0

Hence proved