Question

if alpha and bita are the roots of the of the quadratic equation 2x^2-6x+a and alpha + bita =12 , find a

Answer 1

Step-by-step explanation:

I'm using m and n for alpha and beta.

2x^2+6x+b=0

2x^2+6x+b=0So m+n=-6/2=-3

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2Now ( m/n)+(n/m)=(m^2+n^2)/mn

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2Now ( m/n)+(n/m)=(m^2+n^2)/mn=((m+n)^2-2mn)/mn

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2Now ( m/n)+(n/m)=(m^2+n^2)/mn=((m+n)^2-2mn)/mn =((-3)^2-2(b/2))/(b/2)

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2Now ( m/n)+(n/m)=(m^2+n^2)/mn=((m+n)^2-2mn)/mn =((-3)^2-2(b/2))/(b/2)=(9-b)/(b/2)

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2Now ( m/n)+(n/m)=(m^2+n^2)/mn=((m+n)^2-2mn)/mn =((-3)^2-2(b/2))/(b/2)=(9-b)/(b/2)=(18-2b)/b

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2Now ( m/n)+(n/m)=(m^2+n^2)/mn=((m+n)^2-2mn)/mn =((-3)^2-2(b/2))/(b/2)=(9-b)/(b/2)=(18-2b)/b=-2+18/b

2x^2+6x+b=0So m+n=-6/2=-3mn=b/2Now ( m/n)+(n/m)=(m^2+n^2)/mn=((m+n)^2-2mn)/mn =((-3)^2-2(b/2))/(b/2)=(9-b)/(b/2)=(18-2b)/b=-2+18/bie always less than 2 (b is less than 0)

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