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Misha Larkins [42]
2 years ago
9

Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1

4 = 0 Pairs {-2, 7} {2, -7} {-2, -7} {7, 2}
Mathematics
2 answers:
Anna [14]2 years ago
5 0

Answer:

{-2,7} -> a^2-5a-14=0

{2,-7} -> a^2+9a+14=0

{-2,-7} -> a^2+5a-14=0

{7,2} -> a^2-9a+14=0

I did the math so these should all be correct :)

sattari [20]2 years ago
3 0
We have that

N 1)
a²<span> − 9a + 14 = 0 
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 9a+20.25)=-14+20.25

Rewrite as perfect squares

(a-4.5)²=6.25--------> (a-4.5)=(+/-)√6.25

a1=4.5+√6.25-----> a1=7

a2=4.5-√6.25-----> a2=2

the solution problem N 1 is the pair {7, 2}


N 2) 

a²<span> + 9a + 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

(a+4.5)²=6.25--------> (a+4.5)=(+/-)√6.25

a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3) 

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

the solution problem N 3 is the pair {2, -5}


N 4)

a²<span> + 5a − 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 5a+6.25) =14+6.25

Rewrite as perfect squares

(a+2.5)² =20.25-------> (a+2.5)=(+/-)√20.25

a1=-2.5+√20.25-----> a1=2

a2=-2.5-√20.25-----> a2=-7

the solution problem N 4 is the pair {2, -7}


N 5) 

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 5a+6.25)=14+6.25

Rewrite as perfect squares

(a-2.5)²=2025--------> (a-2.5)=(+/-)√20.25

a1=2.5+√20.25-----> a1=7

a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

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What are the solutions to the following system?
bija089 [108]

The solutions are (\sqrt{2},-1) \text { and }(-\sqrt{2},-1).

Solution:

Given equation:

-2 x^{2}+y=-5

Add 2 x^{2} on both sides.

-2 x^{2}+y+2 x^{2}=-5+2 x^{2}

y=-5+2 x^{2} -------- (1)

y=-3 x^{2}+5 (given)  -------- (2)

Equate (1) and (2).

-5+2 x^{2}=-3 x^{2}+5

Add 3 x^{2} on both sides.

-5+2 x^{2}+3 x^{2}=-3 x^{2}+5 +3 x^{2}

-5+5x^{2}=5

Add 5 on both sides.

-5+5x^{2}+5=5+5

5x^{2}=10

Divide by 5 on both sides, we get

x^{2}=2

Taking square root on both sides, we get

x=\sqrt{2}, x=-\sqrt{2}

Substitute x=\sqrt{2} in (1).

y=-5+2(\sqrt{2})^2

y=-5+4

y=-1

Substitute x=-\sqrt{2} in (1).

y=-5+2(-\sqrt{2})^2

y=-5+4

y=-1

Therefore the solutions are (\sqrt{2},-1) \text { and }(-\sqrt{2},-1).

Option C is the correct answer.

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musickatia [10]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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