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

sum of roots=a+b=-a
hence b=-2a.........(1)
product of roots=ab=b
canceling b b/s
a=1
put a=1 in equation (1)
b=-2(1)
b=-2

Answer 2

The quadratic formula says the roots are
\[x_1,x_2=\frac{-a\pm\sqrt{a^2-4b}}{2}\]
or
\[\large\begin{cases}
a=\dfrac{-a+\sqrt{a^2-4b}}{2}\\\\
b=\dfrac{-a-\sqrt{a^2-4b}}{2}
\end{cases}\]
The first equation gives
\[3a=\sqrt{a^2-4b}~~\iff~~9a^2=a^2-4b~~\iff~~a^2=-\frac{b}{2}\]
which means \(b<0\), otherwise \(a\) is complex.

Substituting into the second equation in the system, we can have
\[\large -2a^2=\frac{-a-\sqrt{a^2-4(-2a^2)}}{2}\]
We should be able to solve for \(a\) now.
\[\begin{align*}-4a^2&=-a-\sqrt{a^2+8a^2}\\
4a^2&=a+3\sqrt{a^2}
\end{align*}\]
Assume \(a>0\) for now, then \(\sqrt{a^2}=|a|=a\), so
\[4a^2=4a\]
which has roots \(a=0\) and \(a=1\).

If we assumed \(a<0\), we'd have
\[4a^2=-2a\]
which has complex solutions for \(a\) (which presumably aren't desirable).

The roots are said to be non-zero, so \(a=1\), which makes \(b=-2\).

Answer 3

Oops, \(4a^2+2a\) doesn't have complex roots, just negative ones. My mistake

Answer 4

hence to find the value of x from x^2+x(1)-2=0
middle term splitting gives
x^2+2x-x-2=0
x(x+2)-(x+2)=0
(x-1)(x+2)=0
x-1=0 & x+2=0
x=+1,-2
hence least value of x=-2

Answer 5

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
b = -2a^2 and b^2 + ab + b = 0----->b=0 or b + a + 1 = 0
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

Answer 6

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..!!

Answer 7

Since a,b are roots of x^2+ax+b
(x-a)(x-b)=x^2+ax+b
x^2-(a+b)x+ab=x^2+ax+b
=> a=-(a+b) => b=-2a...i and =>ab=b => b(a-1)=0...ii

from i and ii,
(-2a)(a-1)=0
either -2a=0 => a=0
or a-1= 0 =>a=1

Now, when a=0 then b=0
and when a=1 then b=-2

SO, two possible solution of a and b are a=0,b=0 and a=1,b=-2

Answer 8

case a=0,b=0
x^2+ax+b=x^2
since x^2>=0 , its least value is 0.

case a=1 and b=-2
x^2+ax+b=x^2+x-2=x^2+2*x*1/2+(1/2)^2-(1/2)^2-2 = (x+1/2)^2 - 9/4
since (x+1/2)^2-9/4 >=-9/4 , its least value is -9/4

Thus, the least value of the given function is -9/4

Answer 9

can someone explain how this is possible that
(x+1/2)^2-9/4 >=-9/4 , its least value is -9/4

Answer 10

Since square of a number is always greater than or equal to 0.
(x+1/2)^2 >= 0
(x+1/2)^2-9/4 >=-9/4

hence its least value is -9/4

Answer 11

to aryandecoolest
if a=-1/2 &b=-1/2,then it should satisfy the values of sum of roots & product of roots
hence a+b= -a & ab= b
but (-1/2)-(1/2)=-1\[\neq1/2\] =-a
& ab=(-1/2)*(-1/2)=(1/4)\[\neq-1/2=b\]
hence a=-1/2 &b=-1/2 must be rejected........

Answer 12

The least value of a quadratic expression is given by\[\dfrac{-D}{4a}\]Where \(D\) is the discriminant. We can also find the values of \(a,b\) by using sum and product formulas.

Answer 13

i am not able to understand all the people are approaching this problem in different ways and giving different solutions for the problem....
which is the correct way........

Answer 14

Correct: all of them.
Easiest: subjective.

Just pick any one of them and enjoy.

Answer 15

if a=1 &b=-2 then putting the value in \[x ^{2}+ax+b=0\] we get,
\[x ^{2}+x(1)-2=0\] & doing middle term splitting we get
\[x^{2}+2x-x-2=0\]
x(x+2)-(x+2)=0
(x-1)(x+2)=0
x=1&x=2 and \[+2\neq-2\] which is the root b=-2 why???
somebody please explain....

Answer 16

very dramatic!!!

Answer 17

funny!!!

Answer 18

You have to put \(a\) for \(x\), not \(a\)

Answer 19

ok i got it........