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kotykmax [81]
3 years ago
5

Solve each quadratic equation by factoring and check

Mathematics
1 answer:
max2010maxim [7]3 years ago
5 0

hello : <span>
<span>the discriminat of each quadratic equation : ax²+bx+c=0 ....(a ≠ 0) is :
Δ = b² -4ac
1 )  Δ > 0  the equation has two reals solutions : x1,2 =  (-b±√Δ)/2a</span></span>

factoring :

<span>ax²+bx+c = a(x-x1)(x-x2)
2 ) Δ = 0 : one solution : x1 = x2 = -b/2a</span>

factoring  : 

<span>ax²+bx+c = a(x-x1)²
3 ) Δ < 0 : no reals solutions... no </span>

<span>factoring </span>
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3 years ago
Select all the systems of equations that have exactly one solution. Group of answer choices {y=3x+1y=-3x+7 ( I know this one is
ololo11 [35]

Answer:

1) Exactly one solution

2) Exactly one solution

3) No solution

4) Infinite many solution

5) No solution

Step-by-step explanation:

As given

1)

y = 3x + 1     ......(1)

y = -3x + 7   ......(2)

Put the value of y in equation (1) , we get

-3x + 7 = 3x + 1

⇒7 - 1 = 3x + 3x

⇒6 = 6x

⇒x = 1

Now as

y = -3x + 7

⇒y = -3(1) + 7 = -3 + 7 = 4

∴ we get

x = 1, y = 4

Unique Solution ( Exactly one solution )

2)

y = 3x+ 1      ......(3)

y = x+1         .......(4)

Put the value of y in equation (3), we get

x + 1 = 3x + 1

⇒1 - 1 = 3x - x

⇒0 = 2x

⇒x = 0

Now, as

y = 3x + 1

⇒y = 3(0) + 1 = 1

∴ we get

x = 0, y = 1

Unique solution ( Exactly one solution )

3)

y = 3x + 1        ........(5)

y = 3x + 7       ........(6)

Put the value of y in equation (5), we get

3x + 7 = 3x + 1

⇒7 - 1 = 3x - 3x

⇒6 = 0

No Solution

4)

x+y=10        ........(7)

2x+2y=20  .........(8)

Divide equation (8) by 2, we get

x + y = 10    .....(9)

⇒y = 10 - x

Put the value of y in equation (7), we get

x + 10 - x = 10

⇒10 = 10

Infinite many solutions

5)

x + y = 10     .......(10)

x + y = 12     .......(11)

⇒y = 12 - x

Put the value of y in equation (10) , we get

x + 12 - x = 10

⇒12 = 10

No solution

6 0
2 years ago
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