Question

Solve by completing the square:
p^2-3p-88=0

Answer 1

do you know what a "perfect square trinomial" is?

Answer 2

This is what I did so far:
p^2-3p-88=0
+88 +88
p^2-3p=88
then
(b/2)^2=(-3/2)^2
Then I got lost

Answer 3

And i remember hearing it, but I forgot what it is

Answer 4

hmmm does \(\large \bf (a\pm b)^2 = a^2\pm 2ab+b^2\) ring a bell?

Answer 5

ummm wow, no not at all :/

Answer 6

heheh... well... that's a "perfect square trinomial"

notice the pattern
1st element squared
middle is "2" times both
2nd element squared

Answer 7

Right, but it says on this equation that I have to solve by completing the square, and that formulas not in the book

Answer 8

well..... keep that in mind for one
because "completing the square" more or less means just that

Answer 9

I'm kinda confued

Answer 10

\( \bf p^2-3p-88\implies p^2-3p=88\implies (p^2-3p+{\color{red}{ \square ?}}^2)=88\)

any ideas what we need there to get a "perfect square trinomial"

Answer 11

I was going to say 88, but I realized you moved that to the other side. Sorry I'm not sure

Answer 12

Would it be 2?

Answer 13

well..... notice in the parentheses
we have 2 terms
the 1st one
and the middle one

notice the middle one
recall that the middle is 2 times both

so...... what do you think would be the missing term, the 2nd one?

Answer 14

You said the middle is 2 times both, so like 3*3 so 9? Sorry if that's wrong

Answer 15

well... let's see the middle term
is 3p <--- we know that much

so we can say that, the 1st term is "p"

so 2 * p * "2nd term" = 3p

if you solve that for "2nd term".... you'd see what that is :)

Answer 16

How do I solve for second term? I know p=1

Answer 17

Would it be 6?

Answer 18

\(\large \bf 2\cdot p\cdot \square =3p\implies \square =?\)

Answer 19

I am not getting any of this. Are you sure this is completing the square, cause I didn't learn this in class?

Answer 20

well...... when they mean "completing the SQUARE"
is referring to a "perfect SQUARE trinomial" as shown above

Answer 21

I don't know how to do that tho. Is 88 suppose to be in the blank?

Answer 22

hmmm do you know how to do linear simplification? that is, solving for a term or a variable?

Answer 23

umm I don't think so. I'm going to put a example up of how they should me how to do it, so hold on

Answer 24

x^2+6x-3=0
+3 +3
x^2+6=3
(b/2)^2= (6/3)^2= 3^2= 9 Add to both sides
x^2+6x+9=3+9
(x+3)^2=12
sqrt(x+3)^2=sqrt(12)
x=3(+-)3.46
x=0.46 or x=-6.46

Answer 25

If you can't help it's fine I give up lol

Answer 26

hmmm

say for if you had ....
2x + 3 = 5

could you solve that for "x"?

Answer 27

2x+3=5
-3 -3
2x=2
x=1

Answer 28

well... that's simplifying by linear simplification

Answer 29

Right...But that's a little simpler then this because you only have one variable not 2

Answer 30

the example you showed... above looks a bit convoluted
let us find the middle term

gimme a sec

Answer 31

\(\bf p^2-3p-88\implies p^2-3p=88\implies (p^2-3p+{\color{red}{ \square ?}}^2)=88
\\ \quad \\
-------------------------------\\
{\color{purple}{ 2\cdot p\cdot \square =3p\implies 2p\square =3p\implies \square =\cfrac{3\cancel{ p }}{2\cancel{ p }}}}\qquad thus\\
-------------------------------\\
\left[p^2-3p+{\color{red}{ \left(\frac{3}{2}\right)}}^2\right]=88\implies \left(p-\frac{3}{2}\right)^2=88\qquad thus
\\ \quad \\
p-\frac{3}{2}=\pm\sqrt{88}\implies p=\pm\sqrt{88}+\frac{3}{2}\)

Answer 32

the 2nd term is 3/2 because
if we make the 1st term "p"
2nd term "3/2"

then the middle term is \(\bf 2\cdot \textit{1st term}\cdot \textit{2nd term}\implies \cancel{ 2 }\cdot p\cdot \frac{3}{\cancel{ 2 }}\implies 3p\)

Answer 33

OMG! I actually get what you did there. Thanks for typing it out like that, it makes more scenes. So is that all you need or is there still more?

Answer 34

that's.... all....once the perfect square is completed
just take square root to both sides
and add the 3/2

Answer 35

hmmmm actiually wait a sec

Answer 36

this is so hard :/ If you can't get it tho don't worry about it. This is the only one on the whole page I didn't get. I appreciate the help

Answer 37

lemme correct myself some

when we are completing the square
we ADDED \(\left(\frac{3}{2}\right)^2\)

but all we're really doing is "borrowing from our good friend Mr Zero" , 0
so if we add \(\left(\frac{3}{2}\right)^2\)
we also have to SUBTRACT \(\left(\frac{3}{2}\right)^2\)

Answer 38

so.... lemme redo that quick

Answer 39

\(\bf p^2-3p-88\implies p^2-3p=88\implies (p^2-3p+{\color{red}{ \square ?}}^2)=88
\\ \quad \\
-------------------------------\\
{\color{purple}{ 2\cdot p\cdot \square =3p\implies 2p\square =3p\implies \square =\cfrac{3\cancel{ p }}{2\cancel{ p }}}}\qquad thus\\
-------------------------------\\
\left[p^2-3p+{\color{red}{ \left(\frac{3}{2}\right)}}^2\right]-{\color{red}{ \left(\frac{3}{2}\right)}}^2=88\implies \left(p-\frac{3}{2}\right)^2=88+{\color{red}{ \left(\frac{3}{2}\right)}}^2\qquad thus
\\ \quad \\
p-\frac{3}{2}=\pm\sqrt{88+\frac{3}{2}}\implies p=\pm\sqrt{88+\frac{3}{2}}+\frac{3}{2}\)

Answer 40

keep in mind, that we're only "borrowing" from Zero, is all we're doing
so whatever amount we ADD, we also have to SUBTRACT it

Answer 41

hmmm I missed the squared part... shoot lemme fix that

Answer 42

\(\bf p^2-3p-88\implies p^2-3p=88\implies (p^2-3p+{\color{red}{ \square ?}}^2)=88
\\ \quad \\
-------------------------------\\
{\color{purple}{ 2\cdot p\cdot \square =3p\implies 2p\square =3p\implies \square =\cfrac{3\cancel{ p }}{2\cancel{ p }}}}\qquad thus\\
-------------------------------\\
\left[p^2-3p+{\color{red}{ \left(\frac{3}{2}\right)}}^2\right]-{\color{red}{ \left(\frac{3}{2}\right)}}^2=88\implies \left(p-\frac{3}{2}\right)^2=88+{\color{red}{ \left(\frac{3}{2}\right)}}^2\qquad thus
\\ \quad \\
p-\frac{3}{2}=\pm\sqrt{88+\left(\frac{3}{2}\right)^2}\implies p=\pm\sqrt{88+\left(\frac{3}{2}\right)}+\frac{3}{2}\)

Answer 43

hmmm \(\bf p^2-3p-88\implies p^2-3p=88\implies (p^2-3p+{\color{red}{ \square ?}}^2)=88
\\ \quad \\
-------------------------------\\
{\color{purple}{ 2\cdot p\cdot \square =3p\implies 2p\square =3p\implies \square =\cfrac{3\cancel{ p }}{2\cancel{ p }}}}\qquad thus\\
-------------------------------\\
\left[p^2-3p+{\color{red}{ \left(\frac{3}{2}\right)}}^2\right]-{\color{red}{ \left(\frac{3}{2}\right)}}^2=88\implies \left(p-\frac{3}{2}\right)^2=88+{\color{red}{ \left(\frac{3}{2}\right)}}^2\qquad thus
\\ \quad \\
p-\frac{3}{2}=\pm\sqrt{88+\left(\frac{3}{2}\right)^2}\implies p=\pm\sqrt{88+\left(\frac{3}{2}\right)^2}+\frac{3}{2}\)

Answer 44

anyhow... had the squred missing from the fraction in the radical

Answer 45

Wow. Thanks again, you helped so much. I'm defiantly going to ask my teacher to go over this tomorrow.

Answer 46

so... simplifying that we could say that

Answer 47

\(\bf p=\pm\sqrt{88+\left(\frac{3}{2}\right)}+\frac{3}{2}\implies p=\pm \sqrt{88+\frac{9}{4}}+\frac{3}{2}
\\ \quad \\
p=\sqrt{\cfrac{361}{4}}+\cfrac{3}{2}\implies p=\pm \cfrac{\sqrt{361}}{\sqrt{4}}+\cfrac{3}{2}\implies p=\pm \cfrac{19}{2}+\cfrac{3}{2}
\\ \quad \\
p=\pm \cfrac{\cancel{ 22 }}{\cancel{ 2 }}\to \pm 11\)

Answer 48

darn... it.. missed the square again anyhow \(\bf p=\pm\sqrt{88+\left(\frac{3}{2}\right)^2}+\frac{3}{2}\implies p=\pm \sqrt{88+\frac{9}{4}}+\frac{3}{2}
\\ \quad \\
p=\sqrt{\cfrac{361}{4}}+\cfrac{3}{2}\implies p=\pm \cfrac{\sqrt{361}}{\sqrt{4}}+\cfrac{3}{2}\implies p=\pm \cfrac{19}{2}+\cfrac{3}{2}
\\ \quad \\
p=\pm \cfrac{\cancel{ 22 }}{\cancel{ 2 }}\to \pm 11\)