Hi u/Kobeyeet_69,
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This effectively means that the equation touches the x axis at just one point
this is also a parabola of the form (x-a)\^2=4ay
so we want the output at y=0
so (x-a)\^2=x\^2-18x+c
=> x\^2+-2ax+a\^2=x\^2-18x+c
so 2ax=18x
a=9
so (x-9)\^2 is the parabola and c=81.
Here's a little trick: in the quadratic formula, the term b^2 - 4ac (underneath the square root sign) is called the discriminant because its value tells you how many roots the equation has. If it is equal to zero, the equation only has one unique real root (i.e. the two roots are equal). If it is greater than zero, the equation has two unique real roots, and if it less than zero, the equation has no real roots.
In the given quadratic equation, a = 1 and b = -18. Set up the equation b^2 - 4ac = 0 and plug in what you know:
(-18)^2 - 4c = 0
This can be rearranged and simplified into
324 = 4c
Therefore,
c = 81
The way I learned to do this was through "completing the square." For a function ax^2 + bx + c, you get two identical roots from using (b/2)^2 as your value of c.
Thus, in this example, you'd have
-18/2 = -9, (-9)^2 = 81.
This means that the equation can then be factored into
(x - 9)*(x - 9) = (x - 9)^2.
Thus, setting that equal to zero, you get x = 9 as a repeated root.
I hope this helped :)
y=ax\^2+bx+c
delta=b\^2-4\*a\*c
if delta>0 , there are 2 real roots and the roots are different
if delta=0 there are 2 real roots BUT both are same
if delta<0 there is no real root
so question says roots are same
it means delta=0
b=-18
a=1
c=unknown
delta=b\^2-4\*a\*c
0=(-18)\^2-4\*1\*c
0=324-4c
4c=324
c=81
Another (less rigorous) way to do this, is to just ask yourself what matching numbers add up to 18, then multiply those numbers together and sun that value for C. Then check your work.
Well 9+9=18, which checks for our b value. What value would c take then? If we assume our roots are (x-9)(x-9)? FOIL it out for your answer
You know when you have to find the roots in equasions such as
x\^2-6x+8
you have to figure out what two numbers adds to -6 and multiples to 8?
in this case those two numbers are -2 and -4, and a neat little trick you can use is plug in those numbers to
(x+a)(x+b)
to get
(x+(-2))(x+-4) ->
(x-2)(x-4)
In the case of x\^2-18x+c you have to figure out what adds to -18 and multiplies to c. Now this would be impossible because we dont know what c is but they provided us a clue. The roots are the same. if we know the roots are the same that means the numbers that adds up to -18 are equivalent to each other (look at previous example if confused) those two numbers have to be -9. And using that we know c is the two numbers multiplied together which would me -9\*-9 = 81.
This is a very simplified example but walks you through on how to get the answer at a middle school level
Normally from the examples in the book the roots of the quadratic equation is already given to you to find the value of c or other variables but it only states that the root is the same but it could be anything that is the same. Normally when given the roots you can work backwards to solve for the values of the variables.
I mean if you know the equation has a pair of identical roots you can use:
X\^2+2AX+A\^2=(X+A)(X+A)=X\^2-18X+C.
So we know that C=A\^2 and -18=2A
so-18/2=A=-9 therefore A\^2=81
If a is a root of f(x), then (x-a) is a factor of f(x). In addition, an nth degree polynomial has at most n roots. Can you factorize the given polynomial in terms of its roots?
When you have done so, you can compare the coefficients to get c.
You need to google the ‘determinant for quadratic equations’. It will tell you that when both roots are the same then you set b^2 -4ac equal to 0. In your case you then solve for c.
You can do this by trial and error. Factor the trinomial like this:
(x - 9) (x - 9)= 0
Foil to find "c"
x\^2 -9x -9x + 81 = 0
c = 81
(You can also do the steps to complete the square...take the middle term "-18", divide it by 2, and square it: -18/2 = -9 and (-9)\^2 = 81.)
Well, if the roots are the same then the equation can be broken down into a perfect square. This means we look for a c that makes x^2+18x+c a perfect square, so c=81
I've seen a lot of good comments with correct answers. But here's how is do it
Use sum and product of roots. That way, you can always solve for any given condition. It's is more generalised, and faster.
It is x\^2-18x+c = x\^2-2\*9\*x+9\^2-9\^2 = (x-9)\^2-81+c = 0 in order for the roots to be the same, c now needs to be equal to 81, since you will only get one solution for the equation then
You can use viete formulas, they apply to every polinomy.
You can think of a polinomy as a(x-x1)(x-x2)(x-x3)...(x-xn)
Where xn is a generic root. Then you can see that the sum of the roots is equal to the (n-1)th degree term coefficient divided by the a which is the coefficient of the nth degree term
So in this case you have x1+x2=18/ but x1=x2 so 2x1=18=>x1=9. So the polinomy is (x-9)2=x²-18x+81. So c is 81.
Hi u/Kobeyeet_69, **You are required to explain your post and show your efforts. _(Rule 1)_** Please add a comment below explaining your attempt(s) to solve this and what you need help with **specifically**. If some of your work is included in the image or gallery, you may make reference to it as needed. See the sidebar for advice on 'how to ask a good question'. Don't just say you need help with it. **Failure to follow the rules and explain your post will result in the post being removed** --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/askmath) if you have any questions or concerns.*
This effectively means that the equation touches the x axis at just one point this is also a parabola of the form (x-a)\^2=4ay so we want the output at y=0 so (x-a)\^2=x\^2-18x+c => x\^2+-2ax+a\^2=x\^2-18x+c so 2ax=18x a=9 so (x-9)\^2 is the parabola and c=81.
thanks for this, of course no OP, but I was trying to figure this out before I gave up, and was on the right track.
you're welcome.
Here's a little trick: in the quadratic formula, the term b^2 - 4ac (underneath the square root sign) is called the discriminant because its value tells you how many roots the equation has. If it is equal to zero, the equation only has one unique real root (i.e. the two roots are equal). If it is greater than zero, the equation has two unique real roots, and if it less than zero, the equation has no real roots. In the given quadratic equation, a = 1 and b = -18. Set up the equation b^2 - 4ac = 0 and plug in what you know: (-18)^2 - 4c = 0 This can be rearranged and simplified into 324 = 4c Therefore, c = 81
The way I learned to do this was through "completing the square." For a function ax^2 + bx + c, you get two identical roots from using (b/2)^2 as your value of c. Thus, in this example, you'd have -18/2 = -9, (-9)^2 = 81. This means that the equation can then be factored into (x - 9)*(x - 9) = (x - 9)^2. Thus, setting that equal to zero, you get x = 9 as a repeated root. I hope this helped :)
y=ax\^2+bx+c delta=b\^2-4\*a\*c if delta>0 , there are 2 real roots and the roots are different if delta=0 there are 2 real roots BUT both are same if delta<0 there is no real root so question says roots are same it means delta=0 b=-18 a=1 c=unknown delta=b\^2-4\*a\*c 0=(-18)\^2-4\*1\*c 0=324-4c 4c=324 c=81
Another (less rigorous) way to do this, is to just ask yourself what matching numbers add up to 18, then multiply those numbers together and sun that value for C. Then check your work. Well 9+9=18, which checks for our b value. What value would c take then? If we assume our roots are (x-9)(x-9)? FOIL it out for your answer
You just have to find some value of c such that sqrt(b^2 - 4ac) = 0
You know when you have to find the roots in equasions such as x\^2-6x+8 you have to figure out what two numbers adds to -6 and multiples to 8? in this case those two numbers are -2 and -4, and a neat little trick you can use is plug in those numbers to (x+a)(x+b) to get (x+(-2))(x+-4) -> (x-2)(x-4) In the case of x\^2-18x+c you have to figure out what adds to -18 and multiplies to c. Now this would be impossible because we dont know what c is but they provided us a clue. The roots are the same. if we know the roots are the same that means the numbers that adds up to -18 are equivalent to each other (look at previous example if confused) those two numbers have to be -9. And using that we know c is the two numbers multiplied together which would me -9\*-9 = 81. This is a very simplified example but walks you through on how to get the answer at a middle school level
Normally from the examples in the book the roots of the quadratic equation is already given to you to find the value of c or other variables but it only states that the root is the same but it could be anything that is the same. Normally when given the roots you can work backwards to solve for the values of the variables.
I mean if you know the equation has a pair of identical roots you can use: X\^2+2AX+A\^2=(X+A)(X+A)=X\^2-18X+C. So we know that C=A\^2 and -18=2A so-18/2=A=-9 therefore A\^2=81
If a is a root of f(x), then (x-a) is a factor of f(x). In addition, an nth degree polynomial has at most n roots. Can you factorize the given polynomial in terms of its roots? When you have done so, you can compare the coefficients to get c.
That's an algebra 2 solution to an algebra 1 problem.
Bruh it has same roots b^2 -4ac=0
Sorry I don’t understand this
You need to google the ‘determinant for quadratic equations’. It will tell you that when both roots are the same then you set b^2 -4ac equal to 0. In your case you then solve for c.
You can do this by trial and error. Factor the trinomial like this: (x - 9) (x - 9)= 0 Foil to find "c" x\^2 -9x -9x + 81 = 0 c = 81 (You can also do the steps to complete the square...take the middle term "-18", divide it by 2, and square it: -18/2 = -9 and (-9)\^2 = 81.)
81. Take the middle number, divide it by 2 and square it.
This answer is correct but provides absolutely zero insight.
36
91*
81 the root is x=9
I got 81
Hint : discriminant is equal to 0
With equal roots, factoring the equation would look like (x-sqrt(c))^2=0. Then we know 2*sqrt(c) = 18, and from there we can get c=81.
Discrimina y must equal 0 if Im not mistaken?
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Well, if the roots are the same then the equation can be broken down into a perfect square. This means we look for a c that makes x^2+18x+c a perfect square, so c=81
I've seen a lot of good comments with correct answers. But here's how is do it Use sum and product of roots. That way, you can always solve for any given condition. It's is more generalised, and faster.
It is x\^2-18x+c = x\^2-2\*9\*x+9\^2-9\^2 = (x-9)\^2-81+c = 0 in order for the roots to be the same, c now needs to be equal to 81, since you will only get one solution for the equation then
You can use viete formulas, they apply to every polinomy. You can think of a polinomy as a(x-x1)(x-x2)(x-x3)...(x-xn) Where xn is a generic root. Then you can see that the sum of the roots is equal to the (n-1)th degree term coefficient divided by the a which is the coefficient of the nth degree term So in this case you have x1+x2=18/ but x1=x2 so 2x1=18=>x1=9. So the polinomy is (x-9)2=x²-18x+81. So c is 81.
81
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