If the quadratic function 3x2 - 7x + R is a perfect square, find R
\(\frac{49}{24}\)
\(\frac{49}{12}\)
\(\frac{49}{13}\)
\(\frac{49}{3}\)
\(\frac{49}{36}\)
Correct answer is B
3x2 - 7x + R. Computing the square, we have
x2 - \(\frac{7}{3}\) = -\(\frac{R}{3}\)
(\(\frac{x}{1} - \frac{7}{6}\))2 = -\(\frac{R}{3}\) + \(\frac{49}{36}\)
\(\frac{-R}{3}\) + \(\frac{49}{36}\) = 0
R = \(\frac{49}{36}\) x \(\frac{3}{1}\)
= \(\frac{49}{12}\)
At what real value of x do the curves whose equations are y = x3 + x and y = x2 + 1 intersect?
-2
2
-1
9
1
Correct answer is E
y = x3 + x and y = x2 + 1
\(\begin{array}{c|c} x & -2 & -1 & 0 & 1 & 2 \\ \hline Y = x^3 + x & -10 & -2 & 0 & 2 & 10 \\ \hline y = x^2 + 1 & 5 & 2 & 1 & 2 & 5\end{array}\)
The curves intersect at x = 1
Factorize abx2 + 8y - 4bx - 2axy
(ax - 4)(bx - 2y)
(ax + b)(x - 8y)
(ax - 2y)(bx - 4)
(bx - 4)(ax - 2y)
(abx - 4)(x - 2y)
Correct answer is A
abx2 + 8y - 4bx - 2axy = (abx2 - 4bx) + (8y - 2axy)
= bx(ax - 4) 2y(ax - 4) 2y(ax - 4)
= (bx - 2y)(ax - 4)
3
-1
2
-3
1
Correct answer is E
32y + 6(3y) = 27
This can be rewritten as (3y)2 + 6(3y) = 27
Let 3y = x
x2 + 6x - 27 = 0
(x + 9)(x - 3) = 0
when x - 3 = 0, x = 3
sub. for x in 3y = x
3y = 3
log33 = y
y = 1
The factors of 9 - (x2 - 3x - 1)2 are
-(x - 4)(x + 1) (x - 1)(x - 2)
(x - 4)(x - 2) (x - 1)(x + 1)
-(x - 2)(x + 1) (x - 2) (x - 1)
(x - 2)(x + 2) (x - 1)(x + 1)
Correct answer is A
9 - (x2 - 3x - 1)2 = [3 - (x2 - 3x - 1)] [3 + (x2 - 3x - 1)]
= (3 - x2 + 3x + 1)(3 + x2 - 3x - 1)
= (4 + 3x - x2)(x2 - 3x + 2)
= (4 - x)(1 + x)(x - 1)(x - 2)
= -(x - 4)(x + 1) (x - 1)(x - 2)