Which of the following equations is quadratic? x(x2 + 1) = 0 5(4x + 2) = 3 (x + 3)(x + 4) = 5
c or b because it doesnt equal with a 0
the question here, James, is whether you have a quadratic equation or not. Doesn't matter if the given equation ends in 0 or not. What exactly is a quadratic equation? Define it; give an example.
5(4x + 2) = 3 Because, Divide both sides by 55 4x+2=35 4x+2=35 2 Subtract 22 from both sides 4x=35−2 4x=35−2 3 Simplify 35−235−2 to −75−75 4x=−75 4x=−75 4 Divide both sides by 44 x=−754 x=−754 5 Simplify 754754 to 75×475×4 x=−75×4 x=−75×4 6 Simplify 5×45×4 to 2020 x=−720
In math, we define a quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic is y = ax^2 + bx + c, where a, b, and c are numbers and a cannot be 0.
So yeah i would say 5(4x + 2) = 3
and a cannot be 0.
forget my answer it is b
Yeah A is incorrect
to many diffrent answers here
the question here, James, is whether you have a quadratic equation or not. Doesn't matter if the given equation ends in 0 or not. What exactly is a quadratic equation? Define it; give an example
B is right but what u think
the question here, James, is whether you have a quadratic equation or not. Doesn't matter if the given equation ends in 0 or not. What exactly is a quadratic equation? Define it; give an example.
Mathmale have a point ^
i already did do that people
So why u think A is incorrect
me
Yeh
i said i didnt think a was correct
In math, we define a quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic is y = ax^2 + bx + c, where a, b, and c are numbers and a cannot be 0.
Ok
Now lets do C
the first equation needs to be re-written: x(x2 + 1) = 0 contains an exponent (2), which is better written as x^2, not as x2. x(x^2+1)=0. Is this a quadratic or not? Reason? I'm more interested in your understanding the math here; you need to be able to explain why a given answer is correct or not.
^
^^
James? Is the first equation a quadratic or is it not? I'm asking you to explain your reasons. I am aware you've addressed this before, but you have not explained your reasoning.
Let's solve your equation step-by-step. x(x2+1)=0 x3+x=0 Step 1: Factor left side of equation. x(x2+1)=0 Step 2: Set factors equal to 0. x=0 or x2+1=0 x=0 Answer: x=0 that's what i got for a
Solution 1 Divide both sides by 55 4x+2=35 4x+2=35 2 Subtract 22 from both sides 4x=35−2 4x=35−2 3 Simplify 35−235−2 to −75−75 4x=−75 4x=−75 4 Divide both sides by 44 x=−754 x=−754 5 Simplify 754754 to 75×475×4 x=−75×4 x=−75×4 6 Simplify 5×45×4 to 2020 x=−720 that's for b
Please: express your exponentiation correctly: x^2: correct ("square of x") x2: wrong x^3 correct ("cube of x") x3: wrong
Solution 1 Expand x2+4x+3x+12=5 x2+4x+3x+12=5 2 Simplify x2+4x+3x+12x2+4x+3x+12 to x2+7x+12x2+7x+12 x2+7x+12=5 x2+7x+12=5 3 Move all terms to one side x2+7x+12−5=0 x2+7x+12−5=0 4 Simplify x2+7x+12−5x2+7x+12−5 to x2+7x+7x2+7x+7 x2+7x+7=0 x2+7x+7=0 5 Apply the Quadratic Formula Given the general form ax2+bx+c=0ax2+bx+c=0, there exists two solutions where: x=−b+b2−4ac√2a,−b−b2−4ac√2ax=−b+b2−4ac2a,−b−b2−4ac2a. In this case, a=1a=1, b=7b=7 and c=7c=7. x=−7+21−−√2,−7−21−−√2 thats what i got for c
@jjcoolja12 : The point of this question is not to solve for x; it's to identify which of the given equations is a quadratic. That's all.
Yeah
i know
Then why "solve" these equations? wouldn't it make more sense to help james answer the question in front of him: "Which of the following equations is quadratic?" ?
i just like solving them it's just something i grew up on is that against the rules
Yes, because it's a distraction. James never did finish this problem, which involved identifying which of the equations is quadratic. Perhaps if you and the rest of us had focused on that simple question, he would have been able to choose the correct response.
dont worry about this question i got the answer thanks though all of yall
@alivejeremy still there
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