Question

Given the quadratic equation x^2+2x-4=0 has root alpha and beeta then find the quadratic equation of 1/2alpha-3,1/2beeta-3

Answer 1

Answer:

x 2 +2x+2=0

Here, a=1,b=2,,c=2

From quadratic formula,

x= 2×1−2± 2 2−4×2×1

⇒x= 2−2±2i

=−1±i

Therefore,

α=−1+i⇒α 2

=(−1+i)

2=−2i

β=−1−i⇒β

2 =(−1−i)

2 =2i

Now,α 15+β 15

=(α 2 )

7 ⋅α+(β 2 )

7β=(−2i) 7

(−1+i)+(2i) 7

(−1−i)

=(−2) 7

(−i)(−1+i)+2 7

(−i)(−1−i)

=2 7

i(−1+i)+2

7 i(1+i)

=2

7 i(−1+i+1+i)

=2

7 i⋅2i=2 8 ⋅(−1)=−256

Step-by-step explanation:

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Answer 2

Answer:

the right answer is-256