I need help on these few questions. I don't really need the answer I just need to be reminded how to do them and I couldn't find it in my class notes.

A sidewalk of uniform width is built on the inside on all sides of a rectangular grass plot measuring 30m by 40m. The resulting grassy area is then only half the original. How wide is the sidewalk?

A of sidewalk will be= 600m

(36-2x)(40-2x)=600
1440-72x-80x+4x(squared)-600=0
4x(squared)-152x+840=0 divide by 4

x(squared)-38x+210

after that I'm stuck

2. Use the discriminant to determine the nature of the roots.

a) 2x(squared)+3x-4=0

b) 9x(squared)+24x+16=0

unsure of what I'm to do

3. If 3x(squared)+kx+2x+k-1=0 is to have two equal roots, then k=?

Don't know what to do on this question either thanks for your help I just want to know how to do them and if the one I tried is even right.

2. 1) I was okay with the "600" for the area of the sidewalk, assuming that the total area was "thirty by forty", or 30×40 = 1200. But then where did the "36" come from, in the first equation?

2) The "discriminant" is whatever you get inside the square root in the Quadratic Formula. So plug a, b, and c into that part of the Formula, and see what you get. The interpret, according to the rules your book gave you. (The interpretation will vary depending upon things like whether or not you've covered complex numbers.)

3) Hint: For the roots to be the same, then the square-root part of the Formula has to be zero.

Eliz.

3. Do you know what the discriminant is?. $b^{2}-4ac$

If it's zero, then the quadratic has two roots of multiplicity two.

Therefore, for 3, you can solve $(2+k)^{2}-4(3)(k-1)=0$

4. 1. I made a typo it's
(30-2x)(40-2x)=600
4x(squared)-140x+600 divide by 4
x(squared)-35x+150
then do I put it into the quadratic formula?

2. a) square root b2-4ac
square root 9-32
square root -23

but you can't have a negative in the root can u?

b) square root 576-576 = square root of 0= 0

what else am I supposed to do for this question?

3. I'm really confused on this one

5. 1) If you mean "4x<sup>2</sup> - 140x + 600 = 0, and then divide by 4", yes, I would agree.

You can solve the equation any way you like. The Quadratic Formula would be one method.

2-a) You ask "you can't have a negative in the root[,] can [you]?" This will, as mentioned, depend upon whether you are considering complex roots or not. What does your book say?

2-b) The exercise instructs you to "Use the discriminant to determine the nature of the roots." So say what kind of roots this equation has. (The particular terminology will depend upon your book.)

3) The "hint" said to that discriminant would have to be zero. So how far have you gotten with that?

Eliz.

6. 1) I have to find x so would I put x(squared)-35x+150 into quadratic formula if I do I get x=5.01

2) The excercise mentions nothing about complex roots and my notes have nothing to do with determining the kinds of roots the only thing even close to do with labeling roots is extraneous roots which has nothing to do with what I'm doing.

3) I don't understand what I'm even supposed to do that hint means nothing to me I don't even know where to start even with that hint.

7. 1) I may be computing things incorrectly, but I'm getting nice neat whole-number answers. Could you please show your steps? Thank you.

2) Okay. If you haven't covered complex numbers, then you aren't supposed to say anything about "complex zeroes". Just follow the examples from your book: what does it say about the roots when the discriminant is negative? (I don't know if your book is looking for a "solutions" sort of answer or a "graphical interpretation" sort of answer.)

3) Does the following help at all?

. . . . .$\large{3x^2\,+\,(k\,+\,2)x\,+\,(k\,-\,1)\,=\,0}$

What are "a", "b", and "c" (from the Quadratic Formula)? Then what is the value (in terms of "k") of the discriminant? Then what does "k" have to be for the discriminant to be equal to zero?

Eliz.

8. 1) I got x=5 using quadratic formula

2) a) irrational root?
b) rational root?

3) what happened to the x after k

would it be 3x(squared)+ (kx+2)x +(k-1)=0

None of this is coming back at all for this question

9. Originally Posted by 4hgurly
1) I got x=5 using quadratic formula
NO! You will get 2 answers, one of which is 5;
you evidently cannot use the quardatic formula correctly;
SHOW YOUR STEPS as stapel asked you to do...then we can explain what you're doing wrong...

10. Originally Posted by Denis
You will get 2 answers, none of which are 5....
I agree on the "two answers" part, but I'm getting "5" as one of them.

Eliz.

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