# Sum of Squares of Roots of quadratic equations.

1. Mar 16, 2012

### the-ever-kid

1. The problem:

Find the value of $\lambda$ for which the sum of squares of the roots of the equation: $2x^2 + (2\lambda +4)x^2-(1+ \lambda) = 0$ has minimum value.

2. Relevant equations

$\alpha + \beta = -b/a$

$\alpha\beta = c/a$

where a,b,c are coefficients of $x^2 , x$ and constant term.

3. The attempt at a solution

what i did was figuring out that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2(\alpha\beta)$ ...... $(i)$

after that i found out $-b/a$ and $c/a$ plugged it in the equation $i$ and got a quadratic in $\lambda$

$\lambda^2+5\lambda+5$......$(ii)$

i then differentiated it to get the minima values of $ii$ i got $f(\lambda)=-5/4, \lambda \longrightarrow -5/2$

my problem is that the sum of squares of roots cannot be negative, can they?

and the solution options are $2,3,-3,-2$

2. Mar 16, 2012

### Mentallic

You actually have quite an intriguing question there. Out of all of these kinds of questions I did in school, none of them involved this kind of tricky situation.

You're right that the sum of two squares cannot be negative (unless you've already learnt about complex numbers and thus are expected to give the answer you have there), but we'll assume you are supposed to be dealing with real roots, which makes this question quite a bit harder.

Ok so we know that $$\lambda ^2+5\lambda +5=f(\lambda)\geq 0$$ because of the reasoning you presented, if we just took $f(\lambda)=0$ then that means the sum of the squares at this value of $\lambda$ will be 0, which means both roots of the original quadratic $$g(x)=2x^2+(2\lambda +4)x-(1+\lambda)$$ would have to be 0 (because you can only get 0 by taking $0^2+0^2=0$) but this is obviously not possible since g(x) is not of the form $y=ax^2$ with other terms that need to be zero ($2\lambda +4=0$ and $1+\lambda=0$ cannot both be true for any $\lambda$).

So what does this mean? It means that at when $f(\lambda)=0$ those values of $\lambda$ will give us a quadratic g(x) that does not cut the x-axis. They're called complex roots (not real) but what we want are real roots, so how can you tell when the roots of a quadratic are real?

3. Mar 16, 2012

### the-ever-kid

then all the possible solutions lead to the fact that the roots of x are complex.........then do we compare the moduli of the roots?

that would probably be really lengthy.....

BTW if the roots were complex then the answer would be the minima right???...... after differentiating...$g(\lambda)$ and equating it to zero?

4. Mar 16, 2012

### Mentallic

No, the roots can be real, that is they can be real for a specific range of $\lambda$.

What exactly are we solving here?

$$\lambda \in ℝ$$
$$\alpha, \beta \in ℝ$$

Or maybe

$$\lambda \in ℝ$$
$$\alpha, \beta \in ℂ$$

??

Maybe you want help solving each case just for the diversity? It really depends on what your class and teacher would expect of you.
By the way, if you want to solve the case where the roots of g(x) are real, thus $\alpha, \beta \in ℝ$ then heed my advice from the previous post, find what the equation to show that the roots are real would be.

5. Mar 16, 2012

### D H

Staff Emeritus

Sure they can if the roots have non-zero imaginary part -- and that is exactly what is happening when λ = -5/2.

6. Mar 16, 2012

### the-ever-kid

HEy thats exactly what i got when i solved it first but it wasnt given in the solutions.......ive clarified that the solution given was wrong thanks......

7. Mar 16, 2012

### D H

Staff Emeritus
I read the problem as 2x2+(2λ+4)x-(1+λ)=0, for which λ = -5/2 minimizes the sums of the squares of the x solutions to this equation.

Re-reading the original post, you have the equation as 2x2+(2λ+4)x2-(1+λ)=0, which is a rather different problem.

So which is it?

8. Mar 16, 2012

### Mentallic

I'm sure it was a typo.

the-ever-kid, I assumed you weren't supposed to be dealing with complex numbers because of what you said

9. Mar 16, 2012

### emailanmol

I have same doubt as DH.
(lambda =m. i am via cell so no symbol for lamba )
But assuming its a typo and 2m+4 is coefficient of x.

You have to solve two equations to get the answer for real roots.

One is discriminant is positive.
And the other you have solved.

Only in the domain of values for m where D is positive, you have to mimimise m^2+5m+5.

This is for real roots.For complex roots the value is what you found just by taking minima of
m^2+5m+5

10. Mar 17, 2012

### the-ever-kid

it was a typo sorry........ and yup i did it the minima way.....long ago...

11. Mar 17, 2012

### emailanmol

I solved it and none of the options match (for both cases , where you want roots to be real or when you want them to be complex)
However , we can assume the closest possible value to be the answer