Help with The Quadratic Function (1 Viewer)

[Snow Pea!]

How do you find the quadratic equation when they only give you the roots of say.. 2 and -5? (This is the stuff which has alpha beta etc. in it)

 

[shafqat]

How do you find the quadratic equation when they only give you the roots of say.. 2 and -5? (This is the stuff which has alpha beta etc. in it)

if they say its monic (coefficient of x^2 is 1), then u write it in the factored form, ie
f(x) = (x-2)(x+5), then expand

if they dont say its monic, u can only find a quadratic with those roots, as there will be infinitely many

 

[Eagles]

what does that mean: 'find the quadratic equation'??

what is the question trying to ask?

 

[Jaydels]

what does that mean: 'find the quadratic equation'??

what is the question trying to ask?

it's asking you to find a quadratic eqn with roots 2 and -5

 

[nick1048]

shafqat answered this question.

 

[Snow Pea!]

The actual Q says: Find the quadratic equation whose roots are.. 2 and -5. I understand how to find the actual roots from an equation but im not sure how to work backwords. (BTW pls explain like u are to an ametuer.. im kinda slow at this topic) lol

 

[shafqat]

The actual Q says: Find the quadratic equation whose roots are.. 2 and -5. I understand how to find the actual roots from an equation but im not sure how to work backwords. (BTW pls explain like u are to an ametuer.. im kinda slow at this topic) lol

if a and b are roots of a monic quadratic, then the quadratic is (x-a)(x-b). thats it. I dont know how to explain any further

 

[FinalFantasy]

as an amateur, juz do it like (x-a)(x-b)
in ur case a=2 and b=-5
so ur equation can be (x-2)(x+5)

 

[acmilan]

Well what do you do to find solutions to quadratic equations? You factorise and equate the factors to 0.

eg.
y = x²-a²
y = (x - a)(x + a)

To find roots, make y = 0

So (x-a)(x+a) = 0, so (x-a) = 0 or (x+a) = 0, hence x = -a or x = a

In this case you were given what x equals, so you do the exact things, except backwards.

So x = -a or x = a

x+a = 0 or x-a = 0

These are your factors, and multiplying them will give the equation

(x+a)(x-a) = x²-a²

 

[ephemeral]

you sure there's no other point? usually in these questions they give the y intercept or something to find an equation, otherwise as shafqat says there are infinite possibilities with y=a(x+5)(x-2) with a affecting dilation etc. Just remember the null factor law thing , it's easy to do it intuitively for basic linear factor questions like this by remembering the root will be the negative of the number in the factor

if you had another point you could just sub that in to get a unique answer.

 

[Snow Pea!]

the answer says... x(squared) +3x - 10 = 0 so if the answer was (x+2)(x-5) how does it expand from there.. ??

 

[shafqat]

the answer says... x(squared) +3x - 10 = 0 so if the answer was (x+2)(x-5) how does it expand from there.. ??

lol binomial expansion. im teaching my brother this.
(x+2)(x-5) = x(x-5) + 2(x-5) [after multiplying out]
= x^2 - 5x + 2x - 10
= x^2 - 3x - 10

 

[Snow Pea!]

ooooo i see.... lol now i get it! Thankyou sooo much!

 

[rama_v]

Alternatively you could use the equation of the form:
x^2 - (sum of roots)x + (product of roots)
= x^2 - (-5 +2)x + (-5x2)
= x^2 + 3x -10

 

[Snow Pea!]

which would be the easiest to rem?? or does it all depend on the person.. lol cos they both seem pretty simple once they are explained.. but i guess everythin does ay?

 

[acmilan]

I would usually use the one that rama_v suggested, thats because it can be applied to a number of different questions