Problem on Quadratic Equation

[Schrodinger's Neurotoxin]

Hello everyone,

I'd like to apologize in advance if this is the wrong forum; I'm new, and I'm not sure if this question would fall under "Intermediate/Advanced Algebra".

So, the question I'm doing is:

For a, b, and h real numbers, show that the roots of (x-a)(x-b) =^2 are always real.

I expanded the above equation to:

x^2 - (a+b)x + ab = h^2
x^2 - (a+b)x + ab - h^2= 0

For the roots to be real, the equation's discriminate should be greater than or equal to 0 (I think). So I did this:

b^2 - 4ac >= 0
let b = -(a+b)
let a = 1
let c = (ab - h^2)

(-(a+b))^2 - 4(1)(ab-h^2) >= 0
a^2 + 2ab + b^2 - 4ab - 4h^2 >= 0
a^2 -2ab + b^2 - 4h^2 >=0

And now I'm stuck, I'm not really sure how to prove that the original equation's roots are always real. Any ideas?

Thanks in advance.

 

[soroban]

Hello, Schrodinger's Neurotoxin!

You lost a minus-sign in there . . .

For real numbers $\displaystyle a,b,h$, show that the roots of. $\displaystyle (x-a)(x-b)\,=\,h^2$ are always real.

I expanded the above equation to: .$\displaystyle x^2 - (a+b)x + ab - h^2\:=\: 0$

For the roots to be real, the equation's discriminant should be greater than or equal to 0.

So I did this: .$\displaystyle [-(a+b)]^2 - 4(1)(ab-h^2)\:\ge\:0$

. . . . . . . $\displaystyle a^2 + 2ab + b^2 - 4ab + 4h^2\:\ge\:0$
. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \uparrow$

. . . . . . . . . $\displaystyle a^2 -2ab + b^2 + 4h^2 \:\ge\:0$

And now I'm stuck.

You have: .$\displaystyle D \:=\: (a-b)^2 + 4h^2$

Note that $\displaystyle (a-b)^2$ is always greater than or equal to zero,
. . . . . . . and $\displaystyle 4h^2$ is always greater than or equal to zero.

Therefore . . .

 

[Schrodinger's Neurotoxin]

Hi soroban,

Sorry about the minus sign, didn't notice it. :???:

So, since both (a - b)^2 and 4h^2 are always greater than or equal to zero, would it mean that the discriminant is always greater than or equal to zero, and therefore its roots are always real?

Thanks for your help.