I don't get a step with this quadratic question

  • #1
Nathi ORea
82
22
Homework Statement
I did the example, and actually got the same answer (0.215), but am a little confused how they did it.

So I did ; 16 - 4 x 5 x -1 = 28.

Then they factorise it into 4 x 7 for some reason.... then I have no idea what wizardry they perform on the next line.
Relevant Equations
Appreciate any help.
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  • #3
Yup. So I used b2 - 4ac to find 28. All good. but I am not sure why they factorised 28?
 
  • #4
Nathi ORea said:
Yup. So I used b2 - 4ac to find 28. All good. but I am not sure why they factorised 28?
Because they know they are about to take the square root and will want to bring the two outside of it. They are being overly descriptive.
 
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  • #5
What rule says you can bring the 2 outside of it?
 
  • #6
##\sqrt{ab}=\sqrt{a} \sqrt{b}##
 
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  • #7
Oh. I see. I haven't got to that section yet. Thank you!
 
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  • #8
Nathi ORea said:
Oh. I see. I haven't got to that section yet. Thank you!
They are probably assuming that you already know this (i.e., there is not a section coming) so you will probably want to review this on your own.
 
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  • #9
Nathi ORea said:
So I did ; 16 - 4 x 5 x -1 = 28.
Typo in your work. The above should be ##16 - 4(3)(-1) = 28##.
Nathi ORea said:
So I used b2 - 4ac to find 28.
You've been a member here for almost three years. Please take the time to learn at least the rudiments of how to write mathematical expressions. At the very least, b^2 - 4ac. Better yet, using LaTeX to write ##b^2 - 4ac##. When you're creating a post, there's a link to our LaTeX tutorial in the lower left corner.
 
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  • #10
As Mark suggested, just wrap the expression b^2-4ac with ## tags for a start.
 
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  • #11
Nathi ORea said:
Oh. I see. I haven't got to that section yet. Thank you!
It's a convention to take any square factors out of a square root. For example, if you ended up with ##\sqrt 4##, that wouldn't be wrong in itself. But, the convention is to write that as ##\sqrt 4 = 2##.

And, if you have ##\sqrt{28}##, then again you are expected to look for square factors. In this case, ##28## has a factor of ##4##. So, you are expected to write ##\sqrt{28} = 2\sqrt 7##.

Note that some mathematicians take this sort of thing very seriously indeed and that leaving something as ##\sqrt{28}## ought to be punishable by a jail term. I don't see that it matters much, but perhaps that's why I'm not a mathematician.
 
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  • #12
PeroK said:
It's a convention to take any square factors out of a square root. For example, if you ended up with ##\sqrt 4##, that wouldn't be wrong in itself. But, the convention is to write that as ##\sqrt 4 = 2##.

And, if you have ##\sqrt{28}##, then again you are expected to look for square factors. In this case, ##28## has a factor of ##4##. So, you are expected to write ##\sqrt{28} = 2\sqrt 7##.

Note that some mathematicians take this sort of thing very seriously indeed and that leaving something as ##\sqrt{28}## ought to be punishable by a jail term. I don't see that it matters much, but perhaps that's why I'm not a mathematician.
Ahhhh.. so that’s why they did it! Thank you very much 🙏
 
  • #13
PeroK said:
It's a convention to take any square factors out of a square root. For example, if you ended up with ##\sqrt 4##, that wouldn't be wrong in itself. But, the convention is to write that as ##\sqrt 4 = 2##.

And, if you have ##\sqrt{28}##, then again you are expected to look for square factors. In this case, ##28## has a factor of ##4##. So, you are expected to write ##\sqrt{28} = 2\sqrt 7##.

Note that some mathematicians take this sort of thing very seriously indeed and that leaving something as ##\sqrt{28}## ought to be punishable by a jail term. I don't see that it matters much, but perhaps that's why I'm not a mathematician.
If one is looking for a numeric result, ie, a decimal answer (such as the 0.215 quoted in the OP), today you simply punch "SQRT 28" into your hand calculator and move on. That's actually fewer keys pressed than "2 times SQRT 7". In the old days, I think algebra lessons stopped at ##2\sqrt 7##. Maybe because evaluating ##\sqrt 7## was an exercise in table lookups, or log tables, or slide rules; and these were topics/skills mastered prior to algebra.
 
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FAQ: I don't get a step with this quadratic question

How do I identify the coefficients in a quadratic equation?

In a quadratic equation of the form ax^2 + bx + c = 0, the coefficients are the numerical values in front of the variables. 'a' is the coefficient of x^2, 'b' is the coefficient of x, and 'c' is the constant term.

How do I factorize a quadratic equation?

To factorize a quadratic equation ax^2 + bx + c, you need to find two numbers that multiply to ac and add to b. Rewrite the middle term using these numbers and then factor by grouping. If the quadratic is factorable, it will be in the form (dx + e)(fx + g) = 0.

How do I use the quadratic formula?

The quadratic formula is used to find the roots of a quadratic equation and is given by x = (-b ± √(b^2 - 4ac)) / (2a). Substitute the coefficients a, b, and c from your equation into this formula to find the solutions for x.

What is the discriminant and how does it help?

The discriminant of a quadratic equation is given by Δ = b^2 - 4ac. It helps determine the nature of the roots. If Δ > 0, there are two distinct real roots. If Δ = 0, there is one real root. If Δ < 0, there are no real roots, but two complex roots.

How do I complete the square for a quadratic equation?

To complete the square, rewrite the quadratic equation in the form ax^2 + bx + c = 0 as a(x^2 + (b/a)x) = -c. Then add and subtract (b/2a)^2 inside the parentheses. Factor the perfect square trinomial and solve for x by isolating the variable.

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