I can't recall how to simplify square roots and am stuck on the following problem:
(-3a)x = √7x+3
x = -1/4
I am trying to solve for a (since this is part of a group of questions that feed of previous answers, I'm absolutely at a stand-still).
Thanks in advance for your assistance!
clarification with parentheses would be good
(-3a)x = √(7x)+3
Right, sorry. I meant to put them in but I'm a bit frazzled. This lot of homework is due tomorrow so I'm kind of everywhere.
(-3a)x = √(7x)+3
divide by -3x and you have
a = (√(7x)+3)/(-3x)
= √(7x)/(-3x) - 1/x
Now assuming that x >= 0 (so √(7x) is real)
= -1/3 √(7/x) - 1/x
An odd sort of problem. As given here, it really has very little to do with simplifying square roots.
Or, given x = -1/4, √(7x) is not even real, so I still suspect trouble.
I wonder if there was a typo in the worksheet. Either way, I'm grateful to you for typing this out for me, I think I get the idea now (which is good, another worksheet has more of the same).
Thanks Steve!
To solve for a in the equation (-3a)x = √7x+3, you need to isolate the variable a on one side of the equation. Here's how you can do it step by step:
Step 1: Start by dividing both sides of the equation by x:
(-3a)x / x = (√7x+3) / x
This simplifies to:
-3a = √7 + 3/x
Step 2: Now, let's isolate a by getting rid of the coefficient (-3):
-3a = √7 + 3/x
Divide both sides of the equation by -3:
a = - (√7 + 3/x) / 3
Step 3: Simplify the expression by rationalizing the denominator. To rationalize the denominator means to eliminate any square roots in the denominator. In this case, we have a fraction with a square root (√7) and a variable in the denominator (3/x).
Multiply the numerator and denominator by the conjugate of the denominator, which is (3/x):
a = - [(√7 + 3/x) / 3] * [x/3]
= - [x(√7 + 3/x) / 3(3/x)]
Simplify further:
a = - [(x√7 + 3) / 9]
So, the solution for a is:
a = - (x√7 + 3) / 9
Now, substitute the value of x = -1/4 into the equation to find the value of a:
a = - ((-1/4)√7 + 3) / 9
Simplify the expression further to get the final value of a.