Real solutions help

[minisue1]

Hi, I have this question, 'Show that (⋋-2)x^2+⋋x +⋋-2=0 has real solutions in x if and only if (3⋋-4)(⋋-4)≤0.'

So, i figure i need to do this by finding the discriminant: the equation becomes (⋋-2)x^2+(⋋)x +(⋋-2)=0,
so ∆= ⋋^2-4(⋋-2)^2
= -3⋋^2+16⋋-16
= -(3⋋-4)(⋋-4) which is the negative version of what I am supposed to prove, i think...

My question is i) am I going the right way about this?
ii) what do I do to remove the negative, and how do I convert this to be ≤0?

HELP PLEASE!!

 

[mmm4444bot]

Show that (⋋-2)x^2+⋋x +⋋-2=0 has real solutions in x if and only if (3⋋-4)(⋋-4)≤0.

So, i figure i need to do this by finding the discriminant

Using the Discriminant is a good approach, but remember that a quadratic polynomial has Real roots when the Discriminant is greater than or equal to zero.

You found the Discriminant to be -(3⋋-4)(⋋-4)

That's correct. Hence, we need -(3⋋-4)(⋋-4) ≥ 0 for the polynomial to have Real roots.

Multiply both sides of that inequality by -1, and you're done! :cool:

Questions?

 

[Deleted member 4993]

Remember to "flip" the inequality - when you are multiplying both sides by a negative number.

 

[minisue1]

Using the Discriminant is a good approach, but remember that a quadratic polynomial has Real roots when the Discriminant is greater than or equal to zero.

You found the Discriminant to be -(3⋋-4)(⋋-4)

That's correct. Hence, we need -(3⋋-4)(⋋-4) ≥ 0 for the polynomial to have Real roots.

So, do all quadratics have real roots? As the original equation =0, do we naturally assume that the real solutions are +ve, and therefore ≥0?
I am a little confused as to the discriminant being +ve or -ve. Being that the formula gave me a -ve answer, wouldn't I assume it to negative??

 

[Deleted member 4993]

Using the Discriminant is a good approach, but remember that a quadratic polynomial has Real roots when the Discriminant is greater than or equal to zero.

You found the Discriminant to be -(3⋋-4)(⋋-4)

That's correct. Hence, we need -(3⋋-4)(⋋-4) ≥ 0 for the polynomial to have Real roots.

So, do all quadratics have real roots? As the original equation =0, do we naturally assume that the real solutions are +ve, and therefore ≥0?
I am a little confused as to the discriminant being +ve or -ve. Being that the formula gave me a -ve answer, wouldn't I assume it to negative ... No...that depends on the magnitude of ⋋??

when ⋋ = 1 → -(3⋋-4)(⋋-4) = -(-1)(-3) = -3

when ⋋ = 2 → -(3⋋-4)(⋋-4) = -(2)(-2) = 4

See the sign of the discriminant changes!

 

[HallsofIvy]

Using the Discriminant is a good approach, but remember that a quadratic polynomial has Real roots when the Discriminant is greater than or equal to zero.

You found the Discriminant to be -(3⋋-4)(⋋-4)

That's correct. Hence, we need -(3⋋-4)(⋋-4) ≥ 0 for the polynomial to have Real roots.

So, do all quadratics have real roots?

No!

As the original equation =0, do we naturally assume that the real solutions are +ve, and therefore ≥0?

No!

I am a little confused as to the discriminant being +ve or -ve. Being that the formula gave me a -ve answer, wouldn't I assume it to negative??

Those are strange questions! Remember the original problem: "'Show that (⋋-2)x^2+⋋x +⋋-2=0 has real solutions in x if and only if (3⋋-4)(⋋-4)≤0" That wouldn't make sense if all quadratics had real roots!

And think about you reason for looking at the discriminant. The quadratic formula says that the quadratic equation \(\displaystyle ax^2+ bx+ c= 0\) has roots \(\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}\). The discriminant is \(\displaystyle b^2- 4ac\), the expression inside the square root. That square root (and so the solutions) will be real if and only if the discriminant is non-negative. If the discriminant is negative, the roots are complex numbers.

You say "Being that the formula gave me a negative answer" but that is not true. The formula gave you a discriminant that depended on \(\displaystyle \lambda\) and whether the answer is positive or negative depends upon \(\displaystyle \lambda\). That is the whole point of this problem! In order that the roots be real the discriminant must be non-negative. And that means that you must have \(\displaystyle -(3\lambda-4)(\lambda-4)\ge 0\). Multiplying both sides by -1 gives \(\displaystyle (3\lambda- 4)(\lambda- 4)\le 0\) (see how the direction of the inequality changed when we multipied by a negative number). Now you should know that "the product of two positive number is positive and the product of two negative numbers is positive". In order that the product be negtive, the two factors must have opposite sign:

Either \(\displaystyle 3\lambda- 4\ge 0\) and \(\displaystyle \lambda- 4\le 0\) or \(\displaystyle 3\lambda- 4\le 0\) and \(\displaystyle \lambda- 4\ge 0\).

 

[JeffM]

So, do all quadratics have real roots? As the original equation =0, do we naturally assume that the real solutions are +ve, and therefore ≥0?
I am a little confused as to the discriminant being +ve or -ve. Being that the formula gave me a -ve answer, wouldn't I assume it to negative??

Every polynomial of odd degree with real coefficients has at least one real root.

Every polynomial of even degree with real coefficients may have no real roots.

Every polynomial of degree n with real coefficients has at most n distinct real roots.

Putting those together for a quadratic means that every quadratic with real coefficients has 0, 1, or 2 distinct real roots.

As for your confusion about the negative. - 3 is a negative number. - a is not necessarily a negative number. - a means that it is a number of the same magnitude but opposite sign as the number a. If a is negative, - a is positive. A minus sign with an expression says nothing about whether the expression is greater or less than zero.