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Solve.$$4 x^{2}+1=3 x^{2}+9$$...

Question

Solve.$$4 x^{2}+1=3 x^{2}+9$$

Solve. $$4 x^{2}+1=3 x^{2}+9$$



Answers

Solve each equation.
$$\frac{4}{2 x^{2}+3 x-9}+\frac{2}{2 x^{2}-x-3}=\frac{3}{x^{2}+4 x+3}$$

Multiply both sides by the power of three over two, and we end up with X minus two. The expensive left side go away equals plus or minus 27. Because remember, it's a score looking at roots, which means they could be positive or negative. Therefore, we now have. We're writing this in terms of acts. X equals two plus or minus 27. X equals two plus 27 gives us our first solution. X equals 29. X equals two minus 27 gets gives us our next solution, which is axe equals negative 25.

If we multiply both sides by the power of three divided by do we end up with X minus two equals 27 giving us the solution of X equals 29.

We have to solve the linear equation. X, Divided by X squared minus nine plus four, divided by X plus three is equal to three, divided by X squared minus nine. So will regroup the terms and bring the similar terms on the same side of the equation in terms of the denominator. So we'll get extradited by X squared minus nine minus three. Divided by X squared, minus nine is equal to minus four, divided by X plus three. So taking the else him on the left hand side as X squared minus nine will get X minus three. Divided by X squared minus nine equal to minus four, divided by X plus tree. Now we'll factory eyes the denominator on the left hand side as X minus three into X plus three from the algebraic identity s squared minus B squared equals a plus B times a minus bi. This will be equal to minus four, divided by X plus tree now X minus three X minus three will get canceled so we'll be left with one divided by X plus. Tree is equal to four divided by X plus tree. Now if you multiply both sides of the equation with X plus tree, they will also get canceled on. Finally, we'll be left with one is equal to four. Rich is no solution to the given equation because the variable does not exist anymore for which the immigration was defined.

For this question, we're going to solve the equation. Nine x to the fourth power plus five X squared minus four equals zero. To do this, what I'm gonna go ahead and do is try to turn it into a regular quadratic someone to turn X to the fourth power into a squared. And I'm gonna turn X squared into a So we're going to get nine a squared plus five a minus four equals zero. Now that we have a regular quadratic weaken, see if this is factory herbal, or we can use the quadratic formula to solve for a doing, a little work off to this side. I do see that this is factory Herbal into a plus one and nine a minus four. So that's the factors of our quadratic here. Now, what we can do is set them both equal to zero to solve. So we're going to get a equals, negative one and a equals four positive for over. No. Now that we have are two answers for a we need to go back and substitute in X squared for a because, remember, X squared equals a So I'm gonna go ahead and plug in X squared for a equals negative one. An X squared equals 4/9. Now that we have that substituted and we can actually use the square root property to solve here. So we get X squared, I'm gonna take the square root of that, which is going to give me X equals we're going to do plus or minus. Now, the square root of negative one is actually just I. So our answers, plus or minus I there and then I'm gonna do the same thing here. The square root of each side civil. Get X equals plus or minus the square to 4/9. Taking both the square root of the numerator and the denominator gives us plus or minus 2/3. So that's our final answer. I negative, I positive 2/3 and negative. 2/3. Those There are four answers and we're done


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