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!
+128408 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
+128408 2. The largest product obtainable is (1998/2 + 1) (1998/2 - 1) = (1000)(998) = 998000
+128408 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
+128408 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
+128408 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
+128408 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
