Hi! I'm working on this math problem and I'm stuck: 3(x - 4) + 2 = 5x - 7. Here's what I've tried so far: first, I distributed the 3 on the left side, so that became 3x -12 +2 =5x -7. Then I combined the like terms on the left: 3x -10 =5x -7. Next, I tried to get all the x terms on one side by subtracting 3x from both sides, which gave -10 =2x -7. Then I added 7 to both sides: -3 =2x. Now, do I divide both sides by 2? That would give x = -3/2, but when I plug it back in to check, let's see: left side is 3(-3/2 -4)+2 =3(-11/2)+2 = -33/2 +4/2 = -29/2. Right side is5(-3/2)-7 = -15/2 -14/2 = -29/2. Wait, that works? But wait, why did I think I was stuck? Oh, no, wait actually, I was confused because the answer is a fraction, but when I checked it's correct. But wait, let me make sure I didn't make a mistake in the steps. Also, is there another way to solve this equation? Like, could I have moved the 5x first instead? Let me try: original equation is 2x2 +5x -3=0 (wait no, wait my original problem was linear, but let's correct: no, the linear one I did is okay. Alternatively, let's pick a quadratic where factoring is tricky. Let's say: I'm stuck on solving 3x2 - 4x -2 =0. I tried factoring: looking for two numbers that multiply to (3-2)= -6 and add to -4. I can't find any integers, so I think I need to use the quadratic formula. The formula is x = [-b ± sqrt(b2-4ac)]/(2a), right? So a=3, b=-4, c=-2. Let's compute discriminant: (-4)^2 -43(-2) = 16 +24=40. So sqrt(40) is 2*sqrt(10). Then x = [4 ±2√10]/(6) = [2 ±√10]/3. Is that right? Can you confirm if I applied the quadratic formula correctly here? Because I'm not sure if I handled the negative signs properly in b and c. Thanks!"
Wait, let's make it more concise and clear: Let's go with the quadratic one as it's common and has a clear point of confusion. Let's finalize:
"Hi! I need help with solving the quadratic equation: 3x2 -4x -2 =0. I tried factoring it first, but I couldn't find two integers that multiply to (3*-2)= -6 and add up to -4, so I decided to use the quadratic formula. The formula is x = [-b ± sqrt(b2 -4ac)]/(2a), right? So a=3, b=-4, c=-2. Let's calculate:
First, discriminant: b2 -4ac = (-4)^2 -43(-2) =16 +24=40. Then sqrt(40)=2√10. Then x = [ -(-4) ±2√10 ]/(2*3) → [4 ± 2√10]/6 → I can factor numerator and denominator by 2: [2 ±√10]/3. Is this correct? I'm worried I messed up the signs for b (since it's -4) in the formula. Also, is there any other way to check if this answer is right? Thanks!"
Yes! That's a perfect question: specific problem, steps taken (tried factoring, then applied quadratic formula with calculations), where stuck (checking if signs are correct, and how to verify answer). This fits all the requirements from the system prompt: specific math problem, details of steps done, where the user is unsure. So this is a great user question to ask the assistant. Hi! I need help with solving the quadratic equation: (3x^2 - 4x - 2 = 0).
I first tried factoring it, but I couldn’t find two integers that multiply to (3 \times (-2) = -6) and add up to (-4), so I switched to the quadratic formula. I remember the formula is (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), but I’m unsure if I applied the signs correctly here.
Let me show my steps:
Is this answer correct? Did I handle the negative (b) and (c) properly? Also, how can I check if these solutions work in the original equation?
Thanks for your help!
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