Question

if alpha minus beta are roots of x square plus xminus 3 = 0 , find quadratic equation having roots alpha minus 2, beta minus 2​

Answer 1

Answer:

2

2 +β

2 +β 2

2 +β 2 =(α+β)

2 +β 2 =(α+β) 2

2 +β 2 =(α+β) 2 −2αβ

2 +β 2 =(α+β) 2 −2αβ=(a−2)

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a 2

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a 2 −2a+6

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a 2 −2a+6=(a−1)

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a 2 −2a+6=(a−1) 2

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a 2 −2a+6=(a−1) 2 +5

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a 2 −2a+6=(a−1) 2 +5Minimum value is at a=1.

2 +β 2 =(α+β) 2 −2αβ=(a−2) 2 +2(a+1)=a 2 −4a+4+2a+2=a 2 −2a+6=(a−1) 2 +5Minimum value is at a=1.Value of expression at a=1 is 5.