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Quadratic Equation Help? 10 points for best answer?

Quadratic Equation Help? 10 points for best answer? Topic: Rewrite my paper free
May 24, 2019 / By Francisca
Question: I have to rewrite quadratic equations in expanded form into factored form. The thing is, I have a few problems that have a (-1x^2) instead of (1x^2). I'm a bit confused... Can anyone help and provide an explanation? e.g. -1x^2-10x-24=0
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Best Answers: Quadratic Equation Help? 10 points for best answer?

Darcey Darcey | 7 days ago
Since you have an equation you multiply EVERY term by -1 (If the =0 were not there then first factor out -1) multiply by -1 and get x^2 + 10x + 24 = 0 (x + 6)(x + 4) = 0 x + 6 =0.......x + 4 = 0 x = -6....x = -4 IF you have any trouble factoring then take free lessons from me. Not bragging but I am EXCELLENT at teaching factoring Mathcerely, Robert Jones "Teacher/Tutor of Fine Students" Moved Florida to France June 2011 but still tutor world-wide (USA , UK , France , Checz Republic ...) free using Skype. If you need me to explain further or any other math problem just let me know how to get in touch. ********* A PRESENT FOR YOU Ask someone to write a number, say a five-digit number. (Can be a 2-digit.....3-digit.....4-digit....ect..... Suppose the number written is 57836 Now, without showing the asker, you write a number on a sheet of paper and keep it folded. You have to write the number by subtracting 2 from the above number and adding 2 in front which will be.... 257834 Next, ask the person to write another five-digit number below his original number. Suppose he writes 37589. So you now have... 57836 37589 Now you write a five-digit number below it in such a way that each digit is 9 minus digit above. You now have... 57836 37589 62410 Ask the person to add one more 5-digit number. If he adds 54732, you add below it 45267. Note that you decided the number by subtracting each of his digit from 9. Thus, you now have..... 57836 37589 62410 54732 45267 Next, ask him to add all the number and he gets 257834 Show him the number which you had written as an answer to this addition earlier in the folded paper and surprise him.
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Darcey Originally Answered: How do you solve a quadratic equation with just three points?
Equation of a quadratic equation is: y = ax^2 +bx +c you have 3 points A(x1, y1) B(x2, y2) C(x3, y3) substitute the 3 points into the equation y1 = ax1^2 +bx1 +c y2 = ax2^2 +bx2 +c y3 = ax3^2 +bx3 +c you have 3 equations with 3 unknowns to solve for a, b and c

Brande Brande
Well, first, to get rid of your problem you can multiply the whole equation by -1 to get x^2+10x+24=0. Then, to factor, you think about what two multiples of the third term add up to the coefficient (the number before the variable) of the second term. This always works when the coefficient of the first number is 1, but if it isn't, it will not. 6 times 4 = 24 and 6+4=10. Therefore, the factored form of this equation is (x+6)(x+4)=0 or (x+4)(x+6)=0, either way is the same.
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Alea Alea
First the quadratic formulation for a given equation ax^2+bx+c=0 is right here: -b±sqrt(b^2+(-4*a*c)) / 2a as a effect enable: a=4 b=6 c-4 utilising your calculator, you will desire to get the two values for x (the effect from the quadratic equation above) to be 0.5 and -2 There you flow!
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