# Thread: Need help on math problems by tommorow

1. ## Need help on math problems by tommorow

1. got on my own
2. (root of x) - 2 = 4th root of x
3. (root of 3-3x) - (root of 3x + 2) = 3
4. write a quad equation with integer coef who roots are
(-1 +- root of -2) / 5
5. got on my own
6. got on my own
7. find all p in f(x) = (p + 1) x^2 + px - 1 so that the quadratic equation's roots are real and unequal
8. find the range of the following quadratic function
f(x) = roof of x(x^2) + roof of 8 (x) -2
almost got but cant get vertex here
9. A park in the shape of a rectangle has dimensions 60m by 100m. If the park contains a rectangular garden enclosed by a concrete walkway, how wide is the walkway if the area of the garden is half the area of the park?
Note : the concrete walkway width is uniform

i tried these and i keep getting the wrong answers
i spent 3 hours trying to figure them out but out of 20 problems, i didnt get these nine
Can you guys give me some hints on how to do these problems?
Thanks

2. ## Problems

Hello mathprob

Welcome to Math Help Forum!
2. (root of x) - 2 = 4th root of x
Let $x = u^4$.Then

$\sqrt{u^4}-2 = \sqrt[4]{u^4}$

$\Rightarrow u^2 - 2 = u$

Solve this quadratic for $u$. Then put $x=u^4$ again at the end.

3. (root of 3-3x) - (root of 3x + 2) = 3
You know that $(a-b)^2 = a^2-2ab+b^2$. So square both sides of

$\sqrt{3-3x} - \sqrt{3x+2} = 3$

to get:

$(3-3x) - 2\sqrt{(3-3x)(3x+2)} + (3x+2) = 3$

$\Rightarrow 2 = 2\sqrt{(3-3x)(3x+2)}$

$\Rightarrow (3-3x)(3x+2) = 1$

4. write a quad equation with integer coef who roots are
(-1 +- root of -2) / 2
Use the quadratic formula. You know that

$\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-1\pm\sqrt{-2}}{2}$

So start with $a = 1, b = 1$ and see what value this gives for $c$. If it's a fraction, multiply both sides of $ax^2+bx+c=0$ by something that will get rid of fractions.
7. find all p in f(x) = (p + 1) x^2 + px - 1 so that the quadratic equation's roots are real and unequal
For real unequal roots, $b^2-4ac>0$

So $p^2+4(p+1)>0$

$\Rightarrow p^2+4p+4>0$

$\Rightarrow (p+2)^2>0$

What is the only value of $p$ for which this is not true?

3. 2. i got 16
3. null set
4. i still cant get, btw i typed in the wrong roots, its divided by 5, not 2
7. i got all reals except p cannot equal -2
8. still cant get
9.cant get

4. Originally Posted by mathprob
i tried these and i keep getting the wrong answers
In future, please show your work, so we can "see" where you're needing help. Thank you!

Originally Posted by mathprob
2. (root of x) - 2 = 4th root of x

2. i got 16
You can check the solution to any "solving" problem by plugging it back into the original exercise. In this case:

. . . . . $\sqrt{16}\, -\, 2\, =\, 4\, -\, 2\, =\, 2$

. . . . . $\sqrt[4]{16}\, =\, \sqrt[4]{2^4}\, =\, 2$

Originally Posted by mathprob
3. (root of 3-3x) - (root of 3x + 2) = 3

3. null set
How did you arrive at your answer? Was the original equation as follows?

. . . . . $\sqrt{3\, -\, 3x}\, -\, \sqrt{3x\, +\, 2}\, =\, 3$

You started the solution by squaring both sides:

. . . . . $(3\, -\, 3x)\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, +\, (3x\, +\, 2)\, =\, 9$

. . . . . $5\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 9$

. . . . . $-2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 4$

Then you squared again:

. . . . . $4(6\, +\, 3x\, -\, 9x^2)\, =\, 16$

. . . . . $6\, +\, 3x\, -\, 9x^2\, =\, 4$

. . . . . $0\, =\, 9x^2\, -\, 3x\, -\, 2$

. . . . . $0\, =\, (3x\, -\, 2)(3x\, +\, 1)$

So x = 2/3 or x = -1/3.

Plugging these back into the original equation gives invalid results: neither 1 - 2 nor 2 - 1 equals 3. So there is "no solution".

Originally Posted by mathprob
4. write a quad equation with integer coef who roots are (-1 +- root of -2) / 5
4. i still cant get, btw i typed in the wrong roots, its divided by 5, not 2
To learn how to find a quadratic by working backwards from its zeroes, try here.

Once you have learned the basic technique, set up the exercise as:

. . . . . $\left(x\, -\, \left(\frac{-1\, -\, \sqrt{-2}}{5}\right)\right)\left(x\, -\, \left(\frac{-1\, +\, \sqrt{-2}}{5}\right)\right)$

Note that, since you must have integer coefficients, you'll need to multiply the product above by "25" to clear the denominator.

Originally Posted by mathprob
7. find all p in f(x) = (p + 1) x^2 + px - 1 so that the quadratic equation's roots are real and unequal
7. i got all reals except p cannot equal -2
From what you've learned about the Quadratic Formula, you know that the only way to get two different real roots is for the value inside the square root to be positive. In other words:

. . . . . $b^2\, -\, 4ac\, =\, p^2\, -\, 4(p\, +\, 1)(-1)\, >\, 0$

So you need to solve the quadratic inequality:

. . . . . $(p\, +\, 2)^2\, >\, 0$

By nature of squaring, (p + 2)^2 will be positive for every non-zero value of p + 2. So the only place the inequality will not hold will be for p + 2 = 0, or p = -2. So your solution is correct.

Originally Posted by mathprob
8. find the range of the following quadratic function
f(x) = roof of x(x^2) + roof of 8 (x) -2
almost got but cant get vertex here
8. still cant get
To learn how to find the vertex of a quadratic, try here. If you're stuck on some other aspect of the exercise, please reply showing your work and reasoning so far.

Originally Posted by mathprob
9. A park in the shape of a rectangle has dimensions 60m by 100m. If the park contains a rectangular garden enclosed by a concrete walkway, how wide is the walkway if the area of the garden is half the area of the park?
Note : the concrete walkway width is uniform
9.cant get
What is the total area? (Hint: Multiply the given length by the given width.)

Draw a rectangle, and label its sides with the given dimensions.

Draw a smaller rectangle inside, and label the gap between the rectangles with "x", being the (unknown) width of the walkway.

What expression would stand for the width of the inner rectangle? What expression would stand for the length of the inner rectangle?

What expression would then stand for the area of the inner rectangle?

If the area of the garden is half of the total park area, what is the garden's area? (Hint: Divide the total area by 2.)

You now have an expression, in terms of "x", for the area of the garden, and a numerical value for the area of the garden. Set these equal, and solve.

If you get stuck, please reply showing how far you have gotten in following the provided steps. Thank you!

5. Originally Posted by stapel
In future, please show your work, so we can "see" where you're needing help. Thank you!

You can check the solution to any "solving" problem by plugging it back into the original exercise. In this case:

. . . . . $\sqrt{16}\, -\, 2\, =\, 4\, -\, 2\, =\, 2$

. . . . . $\sqrt[4]{16}\, =\, \sqrt[4]{2^4}\, =\, 2$

How did you arrive at your answer? Was the original equation as follows?

. . . . . $\sqrt{3\, -\, 3x}\, -\, \sqrt{3x\, +\, 2}\, =\, 3$

You started the solution by squaring both sides:

. . . . . $(3\, -\, 3x)\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, +\, (3x\, +\, 2)\, =\, 9$

. . . . . $5\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 9$

. . . . . $-2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 4$

Then you squared again:

. . . . . $4(6\, +\, 3x\, -\, 9x^2)\, =\, 16$

. . . . . $6\, +\, 3x\, -\, 9x^2\, =\, 4$

. . . . . $0\, =\, 9x^2\, -\, 3x\, -\, 2$

. . . . . $0\, =\, (3x\, -\, 2)(3x\, +\, 1)$

So x = 2/3 or x = -1/3.

Plugging these back into the original equation gives invalid results: neither 1 - 2 nor 2 - 1 equals 3. So there is "no solution".

To learn how to find a quadratic by working backwards from its zeroes, try here.

Once you have learned the basic technique, set up the exercise as:

. . . . . $\left(x\, -\, \left(\frac{-1\, -\, \sqrt{-2}}{5}\right)\right)\left(x\, -\, \left(\frac{-1\, +\, \sqrt{-2}}{5}\right)\right)$

Note that, since you must have integer coefficients, you'll need to multiply the product above by "25" to clear the denominator.

From what you've learned about the Quadratic Formula, you know that the only way to get two different real roots is for the value inside the square root to be positive. In other words:

. . . . . $b^2\, -\, 4ac\, =\, p^2\, -\, 4(p\, +\, 1)(-1)\, >\, 0$

So you need to solve the quadratic inequality:

. . . . . $(p\, +\, 2)^2\, >\, 0$

By nature of squaring, (p + 2)^2 will be positive for every non-zero value of p + 2. So the only place the inequality will not hold will be for p + 2 = 0, or p = -2. So your solution is correct.

To learn how to find the vertex of a quadratic, try here. If you're stuck on some other aspect of the exercise, please reply showing your work and reasoning so far.

What is the total area? (Hint: Multiply the given length by the given width.)

Draw a rectangle, and label its sides with the given dimensions.

Draw a smaller rectangle inside, and label the gap between the rectangles with "x", being the (unknown) width of the walkway.

What expression would stand for the width of the inner rectangle? What expression would stand for the length of the inner rectangle?

What expression would then stand for the area of the inner rectangle?

If the area of the garden is half of the total park area, what is the garden's area? (Hint: Divide the total area by 2.)

You now have an expression, in terms of "x", for the area of the garden, and a numerical value for the area of the garden. Set these equal, and solve.

If you get stuck, please reply showing how far you have gotten in following the provided steps. Thank you!
thanks a lot
Here is the work i have
2. works
3. no solutions, which got to
4. 25x^2 + 10x + 3 = y.
7. P > -2
8. Y >= -2 - root of 2
9. since the walkway is uniform i for 100 - 2x, 60 - 2x
so, 4x^2 - 320x + 3000
used quadratic formula and x has 2 solutions but only one works
i got x = 10.85meters

i need to confirm 4 and 9