Help Quadratic Function!!! (1 Viewer)

[renaoak17]

ok two main issues i am having

q1 find the quadratic equation whose roots are 2 and -5

q2 find the value of m in x^2 +2mx - 6 = 0

i really need help ;-(

please show working or and or an explanation on how and why
thnx
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[addikaye03]

ok two main issues i am having

q1 find the quadratic equation whose roots are 2 and -5

q2 find the value of m in x^2 +2mx - 6 = 0

i really need help ;-(

please show working or and or an explanation on how and why
thnx
<!-- / message -->

Q1. There are two methods, one using the relationship between coefficients and one using the definition of roots.

Method 1( R between C)

from (x-alpha)(x-beta)=0.... when expanded we get :
x^2-(alpha + beta)x+(alpha x beta)
so alpha + beta ( which are the roots of the equation, since given)=2+(-5)=-3
alpha x beta= 2(-5)=-10

by subbing into the expression we get x^2+3x-10 as the equation

Method 2 (definition) Since we know that roots occur when y=0, x=2 and x=-5 respectively. therefore x-2=0 and x+5=0
(x-2)(x+5)=0 when expanded x^2+3x-10

 

[addikaye03]

ok two main issues i am having

q1 find the quadratic equation whose roots are 2 and -5

q2 find the value of m in x^2 +2mx - 6 = 0

i really need help ;-(

please show working or and or an explanation on how and why
thnx
<!-- / message -->

is the 2nd one typed up correctly?.. i cant seem to see hoe there could be an answer, It might be my post-maths ability though (exponential decay) lol

 

[gurmies]

Second question, you have to state how many solutions you want there to be...i.e discriminant theory

 

[minijumbuk]

ok two main issues i am having

q1 find the quadratic equation whose roots are 2 and -5

q2 find the value of m in x^2 +2mx - 6 = 0

i really need help ;-(

please show working or and or an explanation on how and why
thnx
<!-- / message -->

1. (x-2)(x+5)=0

or

-(x-2)(x+5)=0

Roots just indicates where the function cuts the x-axis. It's obvious that if the roots are 2 and -5, the equation would be (x-2)(x+5) or -(x-2)(x+5), since subbing in the value of 2 or -5 into either equation would give zero (ie. zero value of y, hence cutting x-axis)

2. m cannot be found. You haven't given any conditions of the quadratic function. It could be any value, as far as this question you've given is concerned. Are there meant to be some extra information, such as "If the function only has 1 root" or "the function has no real roots"?

 

[addikaye03]

Second question, you have to state how many solutions you want there to be...i.e discriminant theory

haha good, i was thinking, have i forgot all the maths i knew already!

 

[AlexJB]

If you just think about q2 for a second you'll realise m can be anything.

x^2 + 2mx - 6 = 0
a=1 b=2m c=-6
Discriminant: b^2 - 4ac = 4(m^2) + 24
In order for roots to occur (aka the original curve to equal 0) discriminant >= 0
4(m^2) + 24 >= 0

True for all values of m (any number, whether it be possible or negative multiplied by itself will give a positive number, except 0. Even if m were 0, 24>=0 is true).

 

[cutemouse]

q1 find the quadratic equation whose roots are 2 and -5
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Sum of the roots = 2 + (-5) = -3/1=-b/a
Product of the roots = 2*-5 = -10/1=c/a

Therefore (a, b, c) = (1, -3, -10)

Now, a quadratic equation is of form ax²-bx+c=0

Therefore equation is x²+3x-10=0

 

[minijumbuk]

People, you are forgetting that there are two answers to the first question: the positive and the negative.

You'll lose marks if you forget the blue one.

Sum of the roots = 2 + (-5) = -3/1=-b/a

Product of the roots = 2*-5 = -10/1=c/a



Therefore (a, b, c) = (1, +3, -10)



Now, a quadratic equation is of form ax²-bx+c=0



Therefore equation is x²+3x-10=0

Case 2: -3/1 = 3/-1
-b/a = 3/-1
b= -3, a = -1, c = -10

 

[Trebla]

The question should be worded find "a" quadratic equation. There are technically infinitely many quadratic equations with those solutions:
e.g. x² + 3x - 10 = 0, - x² - 3x + 10 = 0, 2x² + 6x - 20 = 0, - 10x² - 30x + 100 = 0 etc

 

[cutemouse]

No, there isn't...

x²+3x-10=0 and -x²-3x+10=0 are the SAME, you get:

x=-5, x=2 and -x=5, -x=-2, which are the SAME.

and my working was correct...

People, you are forgetting that there are two answers to the first question: the positive and the negative.

You'll lose marks if you forget the blue one.

Case 2: -3/1 = 3/-1
-b/a = 3/-1
b= -3, a = -1, c = -10

 

[tommykins]

Trebla is correct - there are an infinite amount of quadratics with the roots 2 and -5.

 

[gurmies]

trying to disprove trebla is just fail.

 

[cutemouse]

... Do you guys bother to read? Do you think there's any reason that I quoted minijumbuk ? .. or did I just do it for fun?

Geez, people today...

 

[tommykins]

x2+3x-10=0 and -x2-3x+10=0 are the SAME, you get:

x=-5, x=2 and -x=5, -x=-2, which are the SAME.

and my working was correct...

How are they the same....

 

[Trebla]

No, there isn't...

x²+3x-10=0 and -x²-3x+10=0 are the SAME, you get:

x=-5, x=2 and -x=5, -x=-2, which are the SAME.

and my working was correct...

They give the same solution set, but they are not the same equations. If the question says to find "a" quadratic equation, either would be correct.

 

[cutemouse]

Yes they would be, which was my point. But having 'a' positive is conventional.

IIRC, even to graph it, you can do it concave up or down and get full marks; but I may be wrong.

 

[Lukybear]

There are infinate amounts of parabolas with roots of 2 and -5

10(x2+3x-10)=0, 20(x2+3x-10)=0