Choose the most appropriate strategy and then use it to solve the quadratic equation.

1 answer:

dsp733 years ago

7 0

(x-8) ^ 2 = 121
(x-8) = + / - root (121)
x = 8 +/- root (121)
The solutions are:
x1 = 8 + root (121)
x2 = 8 - root (121)
2a ^ 2 = 8a-6
2a ^ 2-8a + 6 = 0
a ^ 2-4a + 3 = 0
(a-1) (a-3) = 0
The solutions are:
a1 = 1
a2 = 3
x ^ 2 + 12x + 36 = 4
x ^ 2 + 12x + 36-4 = 0
x ^ 2 + 12x + 32 = 0
(x + 4) (x + 8) = 0
The solutions are:
x1 = -8
x2 = -4
x ^ 2-x + 30 = 0
x = (- b +/- root (b ^ 2 - 4 * a * c)) / 2 * a
x = (- (- 1) +/- root ((- 1) ^ 2 - 4 * (1) * (30))) / 2 * (1)
x = (1 +/- root (1 - 120))) / 2
x = (1 +/- root (-119))) / 2
x = (1 +/- root (119) * i)) / 2
The solutions are:
x1 = (1 + root (119) * i)) / 2
x2 = (1 - root (119) * i)) / 2

You might be interested in

at a restaurant the ratio of kids meal, how manys sold to adult meals sold was 9:2. if there were 81 kids meals sold, how many a

DedPeter [7]

9/2 = 81/a....9 kids meals to 2 adult meals = 81 kids meals to a adult meals
cross multiply
9a = 2(81)
9a = 162
a = 162/9
a = 18...so there were 18 adult meals
ratio is : 81/18

6 0

3 years ago

The expression 2.5a+2b gives the cost (in dollars) for a DVDs and b CDs at a moving sale. a. The cost for 1 DVD is $ and the cos

Mekhanik [1.2K]

Answer/Step-by-step Explanation:

Given:

a = no. of DVDs

b = no. of CDs

Total cost ($) for DVDs and CDs = $2.5a + 2b$

a. The cost for 1 DVD = 2.5(a) = 2.5(1) = $2.5

The cost for 1 CD = 2(b) = 2(1) = $2

b. Total cost for 4 DVDs and 4 CDs:

Substitute a = 4, and b = 4 into the equation.

$2.5(4) + 2(4)$

$10 + 8 = 18$

Total cost for 4 DVDs and 4 CDs = $18

c. To find out if $20 would be enough to buy 6 DVDs and 3 CDs, substitute a = 6, and b = 3 into the equation. Solve for the total cost to see if it equals $20 or is less than $20. If it's greater than $20, then it won't be enough.

$2.5(6) + 2(3)$