96.8k views
4 votes
Which of the following are solutions to the equation

Which of the following are solutions to the equation-example-1
User Greggles
by
8.3k points

1 Answer

1 vote

Hello! Your answer would be A and B, 5 and 4.

----

First, move all values to one side. This will give us the quadratic equation we are looking for to solve (and make it much easier).

x² - 3x + 27 = 6x + 7

x² - 9x + 20 = 0

Now, factor the equation. Essentially, turn it into the form of (x + a)(x + b), where a + b must equal -9 in this case (the coefficient of x) and a * b must equal, in this case, 20.

The numbers which when added equal -9 and multiplied equal 20 are negative 5 and negative 4.

Substitute these into (x + a)(x + b).

(x - 5)(x - 4) = 0

Now, recall that anything multiplied by 0 is equal to 0. This can be applied in this scenario because we are looking to make either (x - 5) or (x - 4) equal to 0, as that would set the entire left side equal to 0, and make our equation correct.

To do this, x would either have to equal 5 or 4. This is because 5 - 5 = 0, and 4 - 4 = 0.

Therefore, your two solutions to the equation are 5 and 4.

----

If you would like confirmation:

5² - 3*5 + 27 = 6*5 + 7

25 - 15 + 27 = 30 + 7

37 = 37

----

4² - 3*4 + 27 = 6*4 + 7

16 - 12 + 27 = 24 + 7

31 = 31

----

Hope this helps!

User Kamran Kausar
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories