Help with a math formula?

Questions
1

Hi Folks,
apologies for what will be a dumb question, but if I wanted to find the value of ‘a’ in the formula:
960 = a(a-2)(a-3)(a-4)
how would I go about that?

Thanks a million to those who can help me!
Mike

2

Hi Mike,

Using WL

Solve[960 == a (a - 2) (a - 3) (a - 4), a] // N
(* {{a -> 8.}, {a -> -3.53019}, {a -> 2.2651 - 5.37232 I}, {a -> 
   2.2651 + 5.37232 I}} *)
3

Hi Rohit,
I greatly, greatly appreciate you taking the time to reply to my question.

I’m still at a bit of a loss however… I can see that in the formula, a = 8… but I can’t see where the other figures come from, eg ({a → -3.53019… etc…), and when you say “Using WL”, I’m not sure what ‘WL’ is…

I know this would be basic, and I am a little embarrassed, but this is eluding me… :frowning: :frowning:
Mike

4

You could get the answer on WolframAlpha. Click on the icon that says “Approximate Forms” and you will see the same answer Rohit gave.

5

Mike,

WL is a commonly used acronym for Wolfram Language.

The equation you are trying to find the roots of is a quartic polynomial. So there are four solutions some of which may be degenerate. If you expand and re-arrange

a^4-9 a^3+26 a^2-24 a-960=0 
eq1 = Expand[a (a - 2) (a - 3) (a - 4)] - 960 == 0
(* -960 - 24 a + 26 a^2 - 9 a^3 + a^4 == 0 *)

Factoring yields one root, a = 8.

Factor@eq1
(* (-8 + a) (120 + 18 a - a^2 + a^3) == 0 *)

The remaining cubic has three roots, one real a = -3.53019 and two complex (which are conjugate) a = 2.2651 - 5.37232 I and a = 2.2651 + 5.37232 I

You can visualize the real roots by plotting the function

Plot[Evaluate@First@eq1, {a, -5, 10}]

2336plot.png
2336plot.png450×279 11.2 KB

If you are interested in just integer or real roots you can add a region specification to Solve

Solve[eq1, a \[Element] Integers]
(* {{a -> 8}} *)

Solve[eq1, a \[Element] Reals] // N
(* {{a -> 8.}, {a -> -3.53019}} *)

Hope that helps.

6

Rohit…
Thanks a million for that!!! It helped me enormously.
Fabulous reply, and I am greatly indebted to you… thanks, and thanks, and thanks! :slight_smile: "-)

regards,
Mike

7

Thanks John… you folk on this forum as just so helpful.

Wish there was a way I could return the favour.

regards,
Mike