Sometimes you'll meet an equation that you think you don't know and you'll start up a conversation. After a few minutes of talking, though, it finally hits you—this is just like a quadratic equation. You feel super embarrassed and you try to act like you knew it the whole time, but the equation already noticed. If you want to avoid this situation from happening to you, pay attention to our tips.
Solve the equation .
Some equations dedicate themselves to a rigorous regimen of diet and exercise. This tends to make them a little bottom heavy. They're fractions.
To solve this kind of equation, the first step is to clear out the fractions by multiplying everything by the least common denominator.
(1 – 2x) + 2(x + 1) = 4(x + 1)(1 – 2x)
1 – 2x + 2x + 2 = 4(x – 2x^{2} + 1 – 2x)
3 = 4x – 8x^{2} + 4 – 8x
8x^{2} + 4x – 1 = 0
Aw, we just wrecked their diet. We didn't think a single fudge brownie was going to ruin their figure like that. Anyway, from this point everything proceeds as for a quadratic equation. Activate the quadratic formula, and
Remember, though, that our original equation had fractions. When looking for solutions, we want to be sure to avoid division by zero. So, we need to make sure that x = -1 and aren't included in our set of solutions. Neither of our answers are one of the forbidden solutions, so we are okay. Now we double-check our answers.
Complete success. Total victory. Correct answer.
Solve the equation x^{4} – 3x^{2} – 4 = 0.
Sometimes an equation doesn't look quadratic, by having some other exponent. However, a little bit of lateral thinking shows that the equation actually has the form a(variable)^{2} + b(variable) + c. For this problem, (variable) = x^{2}. Let's give (variable) a new name, u. We can rewrite the equation as
(x^{2})^{2} – 3(x^{2}) – 4 = 0
u^{2} – 3u – 4 = 0
This is clearly quadratic. And factorable.
(u – 4)(u + 1) = 0
However, the answer is not u = 4 and u = -1. Our answer has to be in terms of x, so let's substitute it back into our answers. u = x^{2} = 4 and u = x^{2} = -1. So, x = ±2 and x = ±i. Unlike equations that start off being quadratic, this particular kind can have more than two solutions. Check it out by substituting them into the original equation (with x, not the one with u).
f(±2) = (±2)^{4} – 3(±2)^{2} – 4 = 16 – 12 – 4 = 0
f(±i) = (±i)^{4} – 3(±i)^{2} – 4 = 1 – 3(-1) – 4 = 0
The real nasty disguises, though, are the ones like x^{2/3} + x^{1/3} + 1, which has u = x^{1/3}. Sometimes it takes a bit of imagination to see how a particular function can be made quadratic. So get creative. Maybe pretend it's a unicorn or something.
Solve for the roots of the following equation. x^{2/3} + x^{1/3} + 1 = 0 |
Solve for the roots of the following equation. 9x^{4} – 4 = 0 |
Solve for the roots of the following equation. |
Solve for the roots of the following equation.
Solve for the roots of the following equation.
x^{6} – 2x^{3} – 8 = 0
Solve for the roots of the following equation.
Solve for the roots of the following equation.
-2x^{3} + 6x^{3/2} – 4 = 0
Solve for the roots of the following equation.
x^{5} + 3x^{3} – 4x = 0