1. ## Re: Factorise Equation

Originally Posted by Prove It
If you check the discriminant (-5)^2 - 4(1)(11) = 25 - 44 = -19, since this is negative that means it does not factorise over the reals. So yes, that's the final answer.
What is meaning by discriminant ??
(-5)^2 - 4(1)(11) = 25 - 44 = -19

where you get all those value ?

2. ## Re: Factorise Equation

Originally Posted by Prove It
It's correct, but it's not complete. You now need to expand everything contained in the square brackets, collect like terms, and see if that factorises as well...
Is there an echo in here?

CB

3. ## Re: Factorise Equation

Originally Posted by gilagila
What is meaning by discriminant ??
(-5)^2 - 4(1)(11) = 25 - 44 = -19

where you get all those value ?
If you have to ask, then it's (probably) not something you've been taught. The last answer you gave a few posts ago is correct. That's it. Finito. The End.

4. ## Re: Factorise Equation

Originally Posted by mr fantastic
If you have to ask, then it's (probably) not something you've been taught. The last answer you gave a few posts ago is correct. That's it. Finito. The End.
I find it hard to believe that they are studying factorising higher polynomials if they haven't covered quadratics...

5. ## Re: Factorise Equation

Originally Posted by Prove It
I find it hard to believe that they are studying factorising higher polynomials if they haven't covered quadratics...
Currently i just applying for some online degree courses, one of the subject is algebra and function. I need to study by myself only, so it's a bit difficult for me when the assignment given before any class started and need to submti in a short time period.

Meaning that may be the question from the assignment is on Topic 5 or Topic 9 but current the class not even start Topic 1.

By the way, here are some continue for the question.

Show that x^2-5x+11 > 0 for all the integers

is it meaning that need to subtitute the x with 1,2,3,4 in the equation , what ever number get the value >0 is countable ?

6. ## Re: Factorise Equation

Originally Posted by gilagila
Currently i just applying for some online degree courses, one of the subject is algebra and function. I need to study by myself only, so it's a bit difficult for me when the assignment given before any class started and need to submti in a short time period.

Meaning that may be the question from the assignment is on Topic 5 or Topic 9 but current the class not even start Topic 1.

By the way, here are some continue for the question.

Show that x^2-5x+11 > 0 for all the integers

is it meaning that need to subtitute the x with 1,2,3,4 in the equation , what ever number get the value >0 is countable ?
Complete the square...

\displaystyle \begin{align*} x^2 - 5x + 11 &= x^2 - 5x + \left(-\frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 + 11 \\ &= \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{44}{4} \\ &= \left(x - \frac{5}{2}\right)^2 + \frac{19}{4} \end{align*}

Since $\displaystyle \left(x - \frac{5}{2}\right)^2 \geq 0$ for all $\displaystyle x$, then $\displaystyle \left(x - \frac{5}{2}\right)^2 + \frac{19}{4} > 0$ for all $\displaystyle x$.

7. ## Re: Factorise Equation

Originally Posted by Prove It
If you check the discriminant (-5)^2 - 4(1)(11) = 25 - 44 = -19, since this is negative that means it does not factorise over the reals. So yes, that's the final answer.
After searching on the internet and text book, found the meaning of Discriminant
b^2 - 4ac = (-5)^2 - 4(1)(11) = 25 - 44 = -19 , if is negative, it does not have a real square root and hence the equation will have no real number roots.

8. ## Re: Factorise Equation

Originally Posted by Prove It
Complete the square...

\displaystyle \begin{align*} x^2 - 5x + 11 &= x^2 - 5x + \left(-\frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 + 11 \\ &= \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{44}{4} \\ &= \left(x - \frac{5}{2}\right)^2 + \frac{19}{4} \end{align*}

Since $\displaystyle \left(x - \frac{5}{2}\right)^2 \geq 0$ for all $\displaystyle x$, then $\displaystyle \left(x - \frac{5}{2}\right)^2 + \frac{19}{4} > 0$ for all $\displaystyle x$.

Just a question, If based on Discriminant, when b^2-4ac is negative, so no real root, b^2 - 4ac < 0
How come we can use (x - 5/2)^2 + 19/4 > 0

2nd question, when we use the formula,
x= -b +- Sq root b^2-4ac / 2a
x= 5 + sq root.-19/2 , 5-sq. root-19/2

If based on the question, Show that x^2 - 5x + 11 > 0 for all integers x., i should use the answer from the formula or based on the answer from the completing the square ?

9. ## Re: Factorise Equation

Again, if we sketch graph g(x) = x^2 - 5x + 11

will be look like this ??

10. ## Re: Factorise Equation

Originally Posted by gilagila

Just a question, If based on Discriminant, when b^2-4ac is negative, so no real root, b^2 - 4ac < 0
How come we can use (x - 5/2)^2 + 19/4 > 0

2nd question, when we use the formula,
x= -b +- Sq root b^2-4ac / 2a
x= 5 + sq root.-19/2 , 5-sq. root-19/2

If based on the question, Show that x^2 - 5x + 11 > 0 for all integers x., i should use the answer from the formula or based on the answer from the completing the square ?
All the discriminant tells you is how many solutions there are to a quadratic equation. If the discriminant is negative and there are no solutions, it's possible that the quadratic is always positive, but it's also possible that the quadratic is always negative...

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