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4 years ago

Sum of root of quadratic equation

Mathematics

1 answer:

Alenkinab [10]4 years ago

8 0

Answer:

-b/a

Step-by-step explanation:

ax2+bx+c = 0- quadratic equation, root equation

to find the sum of the roots, turn the 2nd coefficient negative and divide it by the first coefficient

-b/a

hope this helps, plz vote my answer the rainiest if it does!

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JUST ANSWER PLEASE!!! QUICK

jeka57 [31]

Answer:

Options 1 and 3

Step-by-step explanation:

We should know that, the system of linear equations can be treated as matrices, i.e: we can modify or make any operations provide that we must apply the same operation for all terms of each equation.

Given: the solution for the following system is (2,9)

Px + Qy = R ⇒(1)

Tx + Uy = V ⇒(2)

We will check which system of equation has the same solution.

System A) Px + Qy = R

(P+T)x + (Q+U)y = R+V ⇒(3)

So, By summing (1) and (2) we will get the equation (3)

So, system A has the same solution (2,9)

System B) Px + Qy = R

(P+2T)x + (Q+2U)y = R-2V ⇒(4)

By multiplying equation (2) by 2 and add with equation (1), we will get:

(P+2T)x + (Q+2U)y = R+2V

Which is not the same as equation (4)

So, system B has not the same solution (2,9)

System C) (T-P)x + (U-Q)y = V-R ⇒(5)

Tx + Uy = V

By multiplying equation (1) by -1 and add with equation (2), we will get the equation (5)

So, system C has the same solution of (2,9)

System D) (T-P)x + (Q+U)y = V-R ⇒(6)

Tx + Uy = V

We cannot get equation (6) by the same operation over equation (1)

Note the coefficient of x and y⇒ (T-P) and (Q+U)

They must be (T+P) and (Q+U) OR (T-P) and (Q-U)

So, system D has not the same solution of (2,9)

System E) (5T-P)x + (5U-Q)y = V-5R ⇒ (6)

Tx + Uy = V

By subtract equation (1) from 5 times equation (2), we will get:

(5T-P)x + (5U-Q)y = 5V-R

Which is not the same as equation (6)

So, system E has not the same solution (2,9)

As a conclusion, the systems which have the same solution are:

Options 1 and 3

5 0

3 years ago

Solve 5r2 – 12 = 68. {±4} {±5} {±3} {±6}

sashaice [31]

Hey there!

Assuming you meant

$5r^2 - 12 = 68$

If so, follow these steps so it can be a little bit easier to solve

Firstly, we have to $add$ by $12$ on each of your sides

$5r^2 -12 +12 \\ \\ 68 + 12$

$5r^2 = 5r^2 \\ \\ 68 + 12 = 80$

We get: $5r^2 =80$

Now, we have $divide$ by $5$ on each of your sides

$\frac{5r^2}{5} = \frac{80}{5}$

$\frac{80}{5} = 16$

We get: $r^2 = 16$

$r =$ ± $\sqrt{16}$

Therefore , your answer is: $r =$ ± $4$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

3 0

3 years ago

Please help me by answering this?

Dmitrij [34]

Answer:

180

Step-by-step explanation:

$\frac{6}{2} x^{2} | \frac{8}{2}$

192-12=180

8 0

3 years ago

Find the area of the shape below

maks197457 [2]

Answer:

204 ft^2

Step-by-step explanation:

triangle on the left:

A= l x w x 1/2

A = 12 x 5 x 1/2

A = 30 ft^2

triangle on the right:

Same thing

A = 30 ft^2

square in the middle:

A = l x w

A = 12 x 12

A = 144 ft^2

Add them up:

30 + 30 + 144 = 204 ft^2

3 0

3 years ago

PYTHAGORAS THEROM AND TRIGONOMETRY RATIO

marysya [2.9K]

Answer:

1) ΔACD is a right triangle at C

=> sin 32° = AC/15

⇔ AC = sin 32°.15 ≈ 7.9 (cm)

2) ΔABC is a right triangle at C, using Pythagoras theorem, we have:

AB² = AC² + BC²

⇔ AB² = 7.9² + 9.7² = 156.5

⇒ AB = 12.5 (cm)

3) ΔABC is a right triangle at C

=> sin ∠BAC = BC/AB

⇔ sin ∠BAC = 9.7/12.5 = 0.776

⇒ ∠BAC ≈ 50.9°

4) ΔACD is a right triangle at C

=> cos 32° = CD/15

⇔ CD = cos32°.15

⇒ CD ≈ 12.72 (cm)

Step-by-step explanation:

7 0

3 years ago

Sum of root of quadratic equation​

Sum of root of quadratic equation