maths help

Question

If a and b (not equal zero) are the roots of the equation x^2+ax+b=0, then find least value of x^2+ax+b (x belongs R)

Answer 1

    If a and b are roots, then when you substitute them in, you get an expression that must be true. So,

a^2 + a^2 + b = 0                      and               b^2 + ab + b = 0

2a^2  + b = 0                                                   b(b + a + 1) = 0

b = -2a^2                                                          b=0 or b + a + 1 = 0

                                                                              ^ 

                                                                          never true from original problem

So, we can substitute the first expression into the second and get -2a^2  + a + 1 = 0. Multiply by -1 and get

2a^2 - a - 1 = 0 

(2a + 1)(a-1) = 0 

a=-1/2 or a=1

 

so a = -1/2 is the least value. Substitute back into first expression to get b=-2(-.5)^2 or b = -1/2 

BUT I'm not sure what you mean by the least value of x^2+ax+b???  

Do you mean find the minimum value of f(x) = x^2 - 1/2x - 1/2, substituting a and b back in? If so, the minimum value is on the axis of symmetry for this parabola. Axis of symmetry formula is x=-b/2a or x= (.5)/2 which is x=1/4. substitute that in and get 

(1/4)^2 - 1/2(1/4) -1/2 which is -9/16. So minimum point has coordinates 

(1/4 , -9/16 )

Hope this helps.