Question

I do not know how to solve for n such that each epression is a oerfect square trinomial please help

Answer 1

If you have an equation like \[y = ax^2+bx+c\]it will be a perfect square if (and only if) \[b^2-4ac = 0\]

If you want more assistance from that, you'll need to put up more information about your problem.

Answer 2

Thanks look it says x^2-14+n and i know i have to put it like this: x^2-14=0 but i do not know whats after that 😖

Answer 3

well, \[x^2-14x+n=0\]that gives you what for \(a,b,c\) if you compare with \[ax^2+bx+c=0\]?

Answer 4

is it \[x^2-14+n\] or \(x^2-14x+n\)? Makes a big difference in the answer

Answer 5

It is x^2-14x+n

Answer 6

Okay, so comparing with my equation, what are the needed values of \(a,b,c\) for the two to be the same?

Answer 7

Please help me look it says find the value of n such that each expression is a perfect square trinomial. And the first problem is x^2-14x+n and my teacher told us to put is like this: x^2-14x=0 but the problem is that i dont know what is the next step

Answer 8

I'm telling you a general way to do it that will work no matter which variable is missing, unlike the way your instructor suggested. However, if you want to do it their way, we can do that. Tell me which you prefer.

Answer 9

Teach me your way please

Answer 10

Let's do both, then you can compare. We'll do the instructor's way first.

\[x^2-14x+n = 0\]So the idea is that a perfect square will be of the form \[(x+a)^2 =(x+a)(x+a) = x^2 + ax + ax + a^2 = x^2 + 2ax + a^2\]

Agreed so far?

Answer 11

Yup

Answer 12

We have \[x^2-14x + n = 0\]\[x^2+2ax+a^2 = 0\]
to make them equivalent, \(-14x = 2ax\) and \(n = a^2\)

We can simplify\[-14x=2ax\]\[-14=2a\]\[-7=a\]

So to make our perfect square, we take half of the coefficient of \(x\), square it, and add it to both sides. We have to add it to both sides to preserve the equality.

Answer 13

OH MY GOSH i understand it now, im struglling a little bit but i think i get you thank you :)

Answer 14

So we would add \((-7)^2=49\) to both sides, and rewrite the left side as a square:
\[x^2-14x +49 = 49\]\[(x-7)^2 = 49\]

Answer 15

Now, my method:

\[x^2-14x + n = 0\]\[ax^2+bx+c=0\]\[a=1,\,b=-14,\,c=n\]For that to be a perfect square, the discriminant of the equation must be 0. The discriminant for a quadratic is given by \[\Delta = b^2-4ac\]
\[(-14)^2-4(1)(n) = 0\]\[196-4n=0\]\[196=4n\]\[n=49\]

I could also solve one where the middle term was missing:

\[x^2+nx +16 = 0\]\[ax^2+bx+c=0\]\[a=1,\,b=n,\,c=16\]\[\Delta = b^2-4ac=n^2-4(1)(16)=0\]\[n^2=64\]\[n=\pm8\]

If you check, both \[x^2-8x+16=0\] and \[x^2+8x+16=0\]are perfect squares:
\[(x+4)^2 = x^2+4x+4x+16 = x^2+8x+16\]\[(x-4)^2 = x^2-4x-4x+16 = x^2-8x+16\]

Answer 16

And I trust you can see that I could also find a missing first term with the same approach, but I'm not going to do an example.

Answer 17

Omg you area genius i love the way you teached me jts so perfect and awesome thank you so much thank you thank you thank you

Answer 18

Hey it might be a little too much now but i have to find the value of n such that each expression is a perfect square trinomial. (Again) And i have to deal with fraction so x^2-4/9x+n

Answer 19

\[x^2-\frac{4}{9}x+n\]is that your equation?

Answer 20

Yes

Answer 21

What is half of \(-\dfrac{4}{9}\)?

Answer 22

hint: divide the numerator by 2 rather than multiplying the denominator by 2

Answer 23

Its. 2.25

Answer 24

no, it's not, and you should do you work in fractions rather than decimals

Answer 25

Its 2/3 i think

Answer 26

\[-\frac{4}{9}*\frac{1}{2} = -\frac{4*1}{9*2} = -\frac{4}{18} = -\frac{2}{9}\]

Squaring that, we get \[-\frac{2}{9}*-\frac{2}{9} = +\frac{2*2}{9*9} = \frac{4}{81}\]

Answer 27

You need to practice working with fractions!

Answer 28

I know its cause im in Algebra 1 and we haven 't reviewed anything for regular math and our STAAR test is coming up and our teacher gont refresh our minds im sorry i know i look dumb but oh well but thank god you tell me to studyy ASAP

Answer 29

Do i always have to find the half of the fraction?

Answer 30

You always take half of the coefficient of the x term, square it, and add to both sides.

Here the coefficient was -4/9, so half of that is -2/9, squared is 4/81

\[(x-\frac{2}{9})(x-\frac{2}{9})= x^2 - \frac{2}{9}x -\frac{2}{9}+\frac{4}{81} = x^2-\frac{4}{9}x + \frac{4}{81}\]