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guapka [62]
3 years ago
8

Find the discriminant, and determine the number of real solutions. then solve 2x^2+3x-9=0

Mathematics
1 answer:
LekaFEV [45]3 years ago
7 0
Discriminant = 
b² - 4ac

3² - 4(2)(- 9)
9 + 72
81

Your equation has two real solutions.

2x² + 3x - 9 = 0
2x² + 6x - 3x - 9 = 0
2x(x + 3) - 3(x + 3) = 0

(2x - 3)(x + 3) = 0

Set each equal to zero:

2x - 3 = 0
2x = 3
x = 3/2

x + 3 = 0
x = - 3

Your two solutions are x = - 3, 3/2
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PolarNik [594]

Answer:

c > 8

where c = cats

Step-by-step explanation:

Note that

> means greater than

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≥ means greater than or equal to  

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7 0
2 years ago
Marianne has been collecting donations for her biscuit stall at the school summer fayre. There are some luxury gift tins of bisc
LenaWriter [7]
<span>Answer:

Number of luxury gift tins of biscuits sold at £5 each: 18
Number of normal packets sold at £1 each: 2
Number of mini- packs of 2 biscuits sold at 10p each: 80

Solution:

Number of luxury gift tins of biscuits sold at £5 each: x
Amount received by the number of luxury gift tins of biscuits sold: £5 x Number of normal packets sold at £1 each: y
Amount received by the number of normal packets sold: £1 y
Number of mini- packs of 2 biscuits sold at 10p each: z
Amount received by the number of 2 biscuists sold: 10p z =£0.1z

She tells Amy that she has received exactly 100 donations in total:
(1) x+y+z=100

A collective value of £100:
Total amount receibed: £5 x + £1 y+ £0.1z= £100 → £ (5x+1y+0.1z)= £100 → 5x+y+0.1z=100 (2)

Her stock of £1 packets is very low compared with the other items:
y<<x
y<<z
Then we can despice the value of "y" and solve for "x" and "z":
y=0:
(1) x+y+z=100→x+0+z=100→x+z=100 (3)
(2) 5x+y+0.1z=100→5x+0+0.1z=100→5x+0.1z=100 (4)

Solving the system of equations (3) and (4) using the substitution method:
Isolating z in equation (3): Subtrating x from both sides of the equation:
(3) x+z-z=100-x→z=100-x

Replacing z by 100-x in the equation (4):
(4) 5x+0.1z=100→5x+0.1(100-x)=100
Eliminating the parentheses:
5x+0.1(100)-0.1x=100→5x+10-0.1x=100
Adding similar terms on the left side of the equation:
4.9x+10=100
Solving for x: Subtrating 10 from both sides of the equation:
4.9x+10-10=100-10→4.9x=90
Dividing both sides of the equation by 4.9:
4.9x/4.9=90/4.9→x=18.36734693

Replacing x=18.36734693 in the equation:
z=100-x→z=100-18.36734693→z=81.63265306

We can round x and z: x=18 and z=81 and solve for y in equations (1) and (2):
(1) x+y+z=100→18+y+81=100
Adding similar terms on the left side of the equation:
99+y=100
Subtracting 99 from both sides of the equation:
99+y-99=100-99→y=1

(2) 5x+y+0.1z=100→5(18)+y+0.1(81)=100→90+y+8.1=100→98.1+y=100→98.1+y-98.1=100-98.1→y=1.9→y=2 different to 1

Suppose y=1. Replacing y=2 in equations (1) and (2):
(1) x+y+z=100→x+1+z=100→x+1+z-1=100-1→x+z=99
(2) 5x+y+0.1z=100→5x+1+0.1z=100→5x+1+0.1z-1=100-1→5x+0.1z=99

Repeating the process:
z=99-x
5x+0.1z=99→5x+0.1(99-x)=99→5x+9.9-0.1x=99→4.9x+9.9=99→4.9x+9.9-9.9=99-9.9→4.9x=89.1→4.9x/4.9=89.1/4.9→x=18.18367346
z=99-18.18367346→z=80.81632653
x=18 and z=80
(1) x+y+z=100→18+y+80=100→y+98=100→y+98-98=100-98→y=2
(2) 5x+y+0.1z=100→5(18)+y+0.1(80)=100→90+y+8=100→y+98=100→y+98-98=100-98→y=2

Answer:
Number of luxury gift tins of biscuits sold at £5 each: 18
Number of normal packets sold at £1 each: 2
Number of mini- packs of 2 biscuits sold at 10p each: 80 </span>
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Answer:

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Step-by-step explanation:

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