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# Need help with 5 questions asap ty

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1545
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1. Expand the product $(t-2)(4t^2 + 16)(t+2)$.

2. Find all values of $x$ that satisfy the equation $\frac {12x}{x^2 + 8} = 2.$

3. Compute the sum of all the solutions of $(3x+1)(x-7)+(x-3)(3x+1)=0$. Express your answer as a fraction.

4. Let $a$ and $b$ be the solutions of the quadratic equation $2x^2 - 8x + 7 = 0$. Find $\frac{1}{2a} + \frac{1}{2b}.$

5. Let $s$ and $t$ be the solutions of the quadratic $4x^2 + 9x - 6 = 0.$ Find $$\frac st + \frac ts.$$

Jan 29, 2018

### 4+0 Answers

#1
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1. Expand the product   $$(t-2)(4t^2 + 16)(t+2)$$

Rewrite   as   ( t - 2) (t + 2)  (4t^2 + 16)    =

(t^2  - 4) * 4 * (t^2 + 4)  =

4* (t^2 - 4) (t^2 + 4)  =

4 (t^4 - 16)  =

4t^4  - 64

2. Find all values of x that satisfy the equation   $$\frac {12x}{x^2 + 8} = 2$$

Multiply  both sides by   x^2 + 8

12x  =  2 (x^2 + 8)       divide both sides by 2

6x  = x^2 + 8            subtract  6x from both sides and rearrange

x^2 - 6x + 8  = 0       factor

(x - 4) (x - 2)  = 0

Setting both factors to 0 and solve for x and we get that

x = 4   and x   = 2

3. Compute the sum of all the solutions of  $$(3x+1)(x-7)+(x-3)(3x+1)=0$$ . Express your answer as a fraction.

We can factor this as    (3x + 1) ( x - 7 + x - 3)  =  0    simplify

(3x + 1)  (2x - 10)  = 0

(3x +1) * 2 * (x - 5)  = 0      divide both sides by 2

(3x + 1)(x - 5)  = 0

Setting each factor to 0 and solving for x gives that   x = -1/3   and x = 5

The sum of these  is   5 - 1/3   =   15/3 - 1/3  =    14 / 3    =  4 + 2/3

EDIT to correct a small typo....thanks to the guest for spotting my error  !!   Jan 29, 2018
edited by CPhill  Jan 29, 2018
#4
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CPhill: I get a slightly different answer on No. 3 as follows:

Solve for x:

(x - 7) (3 x + 1) + (x - 3) (3 x + 1) = 0

Expand out terms of the left hand side:

6 x^2 - 28 x - 10 = 0

The left hand side factors into a product with three terms:

2 (x - 5) (3 x + 1) = 0

Divide both sides by 2:

(x - 5) (3 x + 1) = 0

Split into two equations:

x - 5 = 0 or 3 x + 1 = 0

Add 5 to both sides:

x = 5 or 3 x + 1 = 0

Subtract 1 from both sides:

x = 5 or 3 x = -1

Divide both sides by 3:

x = 5          or          x = -1/3. So, the sum is: 5 - 1/3 =4 2/3

Guest Jan 29, 2018
#2
+2

4. Let a and b be the solutions of the quadratic equation 2x^2 - 8x + 7 = 0

Find     1 / [2a]  + 1 / [2b]    =  (1/2) (1/a + 1/b)  = (1/2) (a + b) / [ab] =

[ a + b ]  / [2ab]

This isn't as hard as it seems

The sum of the roots   =  - [-8 ]  / 2   =  4

So   a + b  =  4

The product of the roots  =  7/2

So   ab   =  7/2     ⇒     2ab  = 7

So

[ a + b ] / [ 2ab]    =       4 / 7   Jan 29, 2018
#3
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5. Let s and t be the solutions of the quadratic 4x^2 + 9x - 6 = 0. Find

s           +        t                         s^2  + t^2

__                ___       =             ________

t                     s                             st

This is much like 4 with a few twists

The sum of the roots  =     -9/4

So   s + t  =  -9/4

Square both sides

s^2 + 2st + t^2  =   81/16     (1)

The product of the roots  =   -6/4   =  -3/2

So  st  = -3/2   ⇒   2st  =  - 3    (2)

Sub  (2) into (1)  and we have that

s^2  - 3  + t^2  =  81/16         add 3 to both sides

s^2 + t^2  =   81/16 + 3

s^2 + t^2  =  81/16 +  48/16

s^2 + t^2  =  129/16

So

s^2 + t^2                     (129/16)              - (129/16) (2/3) =  - (129/3) (2/16)  =

_________     =          _________  =

st                              - (3/2)

- 43 ( 1/ 8)  =

-43 / 8   Jan 29, 2018