pre/cal/trig

Question

 9) 8^x [4^(2x-1)] = √2 

10) log16 ( √3X +5) = 1/4 

11) log (2x-1) - log[(x^2)-9] = -1 

12) |6x+ 13| > 5 

13)x^2 - 24x - 81 =0 

14) 3x^2 = 5x +8 

15) x^4 - x^3 - 19x^2 + 33x + 18 = 0, given 3 is a double root 

16) x(x+3)^2(x-2) ≥ 0

 

Answer 1

9) The exponential equation

Rewrite

Apply power of a power property

Apply product of powers property

Since the bases are common then equate the powers.

Solution .

10) The logarthmic equation

Convert logarithmic term to an exponential term with base 16 and exponent 1/4.

Solution .

Answer 2

11)The logarthmic equation is

Apply quotient rule

Compare it to

Roots are

Solution

12) The absolute inequality is | 6x + 13| > 5

Apply the formula | x | > a then x > a or x < -a

First case : 6x + 13 > 5

6x > 5 -13

6x > - 8

x > -8/6

x > - 4/ 3

Second case : 6x +13 < - 5

6x < - 5 - 13

6x < - 18

x < - 18/6

x < - 3

Solution set

13) The quadratic equation is

Solutions are x = 27 and x = - 3.

Answer 3

14) The equation 3x²  - 5x + 8 = 0

Compare it to

15) The polynomial function  x⁴  - x³  -19x²  + 33x + 18 = 0

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nxⁿ + a_n – 1x^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

If p/q is a rational zero, then p is a factor of 18 and q is a factor of 1.

The possible values of p are   ± 1,  ± 2, ± 3, ± 6, ± 9  and ± 18.

The possible values for q are ± 1

By the Rational Roots Theorem, the only possible rational roots are, p/q =   ± 1,  ± 2, ± 3, ± 6, ± 9  and ± 18.

Make a table for the synthetic division and test possible real zeros.

p/q	1	-1	-19	33	18
1	1	0	-19	14	32
-1	1	-2	-17	50	-32
3	1	2	-13	-6	0

Since f(3) =  0, x  = 3 is a zero. The depressed polynomial is  x³ + 2x² - 13x - 6 = 0.

Answer 4

Contd..

If p/q is a rational zero, then p is a factor of 6 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,  ±2 and  ± 3.

Make a table for the synthetic division and test possible real zeros.

p/q	1	2	-13	-6
1	1	3	-10	-16
-1	1	1	-14	8
-1	1	1	-3	6
3	1	5	2	0

Since f(3)  =  0,   x = 3 is a zero. The depressed polynomial is  x² + 5x + 2 = 0

Solve x² + 5x + 2 = 0

Compare it to

Thefore, zeros are x = 3, 3, - 0.4385 and -4.5615.

Answer 5

16) The polynomial inequality is x(x + 3)²  (x -2) ≥ 0

The factors give us zeros of the polynomial.

Therefore,  zeros are x = 0 , x = - 3  and x = 2

And the  zero values and at the solution to separate the number line into intervals.

Now test  sample values in each interval to determine whether values in the interval satisify the inequality.

Test interval     x - value      Inequality                                    Conclusion

(-∞,-3)                x =  - 4              True

(-3, 0)                 x = - 2                 True

(0, 2)                  x = 1                       False

(2, ∞)                 x = 3                        True

Note that the original inequality contains a “≥ ” symbol, We inlude it into set of solutions at x = -3,0 and 2.

At x = -3

At x = 0

At x = 2

Above three statements are true.

Solution set is (-∞, - 3] U [2, ∞).