We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
59
9
avatar

Hello, I have a few Algebra questions that I need help with.

 

1. Let \(a\) and \(b\) be the solutions of the equation \(2x^2+6x-14=0\). What is the value of \((2a-3)(4b-6)\)?

 

2. What is the greatest product obtainable from two integers whose sum is 1998?

 

3. The product of three even consecutive positive integers is twenty times their sum. What is the sum of the three integers?

 

4. The red parabola shown is the graph of the equation \(x=ay^2+by+c\). Find \(c\).

 

5. Find all numbers \(a\) for which the graph of \(y=x^2+a\) and the graph of \(y=ax \) intersect. Express your answer in interval notation.

 

6. If \(\displaystyle{f(x)=x^{(x+1)}(x+2)^{(x+3)}}\), then find the value of \(f(0)+f(-1)+f(-2)+f(-3)\).

 

7. Suppose that \(f(x)\) and \(g(x)\) are functions on \(\mathbb{R}\) such that the range of \(f\) is \([-5,3]\), and the range of \(g\) is \([-2,1]\). The range of \(f(x)\cdot{g(x)}\) is \([a,b]\). What is the largest possible value of \(b\)?

 

8. \(k, a_2, a_3\) and \(k, b_2, b_3\) are both nonconstant geometric sequences with different common ratios. We have \(a_3-b_3=3(a_2-b_2).\)

Find the sum of the common ratios of the two sequences.

 

9. Given that \(x \) is real and \(x^3+\frac{1}{x^3}=52\), find \(x+\frac{1}{x}\).

 

10. If \(\&x\) is defined as \(\&x = x + 5\) and \(\#x\) is defined as \(\#x=x^2\) , what is the value of \(\#(\&4)\)?

 

 

Thank you!

 Aug 27, 2019
 #1
avatar+103148 
+1

Here's a few

 

1.  2x^2 + 6x - 14  =  0   divide through by  2

 

x^2  + 3x  - 7  = 0

 

The sum of the roots  =  a + b  = -3/1  = -3

The product of the roots =  ab  = (-7)/1  = -7

 

And we have

 

(2a - 3) (4b - 6) =  12ab -12b - 12a + 24  =   12ab - 12 (a + b) + 24   =  12(-7) - 12(-3) + 24 =

 

-84 + 36 + 24  =

 

-84 + 60  =

 

-24

 

 

 

cool cool cool

 Aug 27, 2019
 #2
avatar+103148 
+1

2.  The largest product obtainable  is    (1998/2 + 1) (1998/2 - 1)  =  (1000)(998) = 998000

 

 

 

cool cool cool

 Aug 27, 2019
 #5
avatar
0

2)

 

1998 / 2 = 999

999^2 =998,001

Guest Aug 27, 2019
 #7
avatar+103148 
0

Thanks guest....I was assuming that the integers were unique....but....if not.....your answer is correct

 

 

cool cool cool

CPhill  Aug 27, 2019
 #3
avatar+103148 
+1

3. The product of three even consecutive positive integers is twenty times their sum. What is the sum of the three integers?

 

Let  the integers be     (n - 2), n (n + 2)

 

So

 

n (n - 2) (n + 2)  =  20 [ (n - 2) + n + (n + 2) ]

 

n (n^2 - 4)  =  20 [ 3n ]

 

n^3 - 4n  = 60n

 

n^3   - 60n - 4n  = 0

 

n^3  - 64n  =  0

 

n ( n^2 - 64)  = 0

 

Either n  =  0  (reject)   or   n^2 - 64  = 0  ⇒    n  = 8

 

So   the three integers  are    6, 8 10

 

The product of these = 480

The sum is 24

And 20 times this sum  = 480

 

 

 

cool cool cool

 Aug 27, 2019
 #4
avatar
0

xxxxxxxxx

 Aug 27, 2019
edited by Guest  Aug 27, 2019
 #6
avatar+103148 
+1

4. The red parabola shown is the graph of the equation  x = ay^2  + by + c   find c

 

Since the point    ( -2,0)  is on the graph, we have that

 

So

 

-2  = a(0)^2 + b(0)  + c

 

-2  =  0   +  0   + c

 

-2  = c 

 

 

 

cool cool cool

 Aug 27, 2019
 #8
avatar+103148 
+1

10.   If &x  is defined as &x = x + 5    and  #x  is defined as #x = x^2       what  is # (&4)

 

&4   = &x  =  x + 5  =   4 + 5  =  9

 

And

 

# (&4)  =  # ( 9) =  # x =  x^2  =  9^2    =  81

 

 

cool cool cool

 Aug 27, 2019
 #9
avatar+103148 
+1

9. Given that x is real  and  x^3  + 1/x^3  =  52  , find x + 1/x

 

(x + 1/x)^3  =  x^3   + 3x^2(1/x) + 3x(1/x)^2 + (1/x)^3      simplify

 

(x + 1/x)^2  = x^3 + 3x + 3/x  + 1/x^3

 

(x + 1/x)^2  =  [ x^3 + 1/x^3 ]  + 3 [ x + 1/x]

 

(x + 1/x)^3  =  52  + 3[ x + 1/x]

 

(x + 1/x)^3   - 3 (x + 1/x) - 52  = 0           let   (x + 1/x)   =   a      ......so we have

 

a^3  - 3a  -  52  =  0

 

By the Rational Roots Theorem, the possible roots are  ±( 1, 2, 4, 13, 26, 52 )

 

Note that   4    produces a solution because

 

4^3  - 3(4)  - 52  =

 

64  - 12  - 52   =  

 

64 - 64   = 0

 

Therefore    a   =   (x + 1/x)   =  4

 

 

 

cool cool cool

 Aug 28, 2019

24 Online Users

avatar