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Topic: **Problem solving by quadratic equations****Question:**
I'm learning solving quadratic equations by factoring can you help me with sum of these problems please & thanks you :)
1. x²+4x-12 = 0
2. x²-4x=21
3. n²-4= -3n
4. 3x²+13x+4=0
5. 3p²+8p+5=0
6. 3r²+10r+9=6
7. 2n²+17n+11=3
Thanks

May 26, 2019 / By Ava

x^2+4x-12=0 factors of twelve that will give you a middle term of 4x are 6 and 2 so factors are (x+6)(x-2) pay attention to signs if you need to solve for x, then set bothe parenthesis to =0 and x=-6, 2 and that's how you should do the rest of the problems ))

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Did you like the answer? We found more questions related to the topic: **Problem solving by quadratic equations**

You should have the following patterns so drilled into your head by countless homework exercises that you recognize them whether the numbers are constant or variables. a^-b^2 = (a-b)(a+b) (a+b)^2 = a^2 + 2ab + b^2 (a-b)^2 = a^2 -2ab + b^2 In high school this is the same as 3 time 7 equals 21 in third grade. If you plan to learn math without doing any F**** homework, now is the time to use the technique you planned to substitue.

1. (x+2)(x-6)=0 x+2=0 or x-6=0 (below) (subtract 2 from both sides) x = -2 or x-6=0 (add six to both sides) x = 6 so, x= -2 or 6 2. (x-2)(x-4)=0 x-2=0 or x-4=0 (same procedure as above) x=2 or x= 4

1. x²+4x-12 = 0 (x-2)(x+6)=0 2. x²-4x=21 0=x^2-4x-21 then, (x+3)(x-7)=0 3. n²-4= -3n 0=n^2+3n-4 then, (n-1)(n+4) 4. 3x²+13x+4=0 (3x+1)(x+4)=0 5. 3p²+8p+5=0 (3p+5)(p+1)=0 6. 3r²+10r+9=6 = 3r^2+10r+3 so, 0=(3r+1)(r+3) 7. 2n²+17n+11=3 =2n^2+17n+8 so, 0=(2n+1)(n+8)

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Im only solving two because this could be your hw so 1. To factor 1st find 2 numbers that when multiplied equal -12. Then find a factor pair that when added equals 4 So the pair would be 6 and -2 The answer would be (x+6)(x-2) 2. First move the 21 to the other side which would make x^2-4x-21 Then find factor pairs that would be -7 and 3 Answer is (x-7)(x+3) for those that have a number in front of x^2 make sure the pairs that one of them is multiplied by the number in front

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2x^2 - x - 21 = 0 (multiply 2 and -21, which is -42; find two integers that multiply to -42 but add to -1, which are 6 and -7) 2x^2 + 6x - 7x - 21 = 0 (group terms into pairs) (2x^2 + 6x) + (-7x - 21) = 0 (factor each group) 2x(x + 3) - 7(x + 3) = 0 (make (2x - 7) a factor and (x + 3) a factor) (2x - 7)(x + 3) = 0 (zero-product property) 2x - 7 = 0 or x + 3 = 0 2x = 7 or x = -3 x = 7/2 or x = -3 <===ANSWER 3x^2 + 14x - 5 = 0 (multiply 3 and -5, which is -15; find two integers that multiply to -15 but add to 14, which are 15 and -1) 3x^2 - x + 15x - 5 = 0 (group into pairs) (3x^2 - x) + (15x - 5) = 0 (factor each group) (x(3x - 1) + 5(3x - 1) = 0 (make (x + 5) a factor and (3x - 1) a factor) (x + 5)(3x - 1) = 0 (zero-product property) x + 5 = 0 or 3x - 1 = 0 x = -5 or 3x = 1 x = -5 or x = 1/3 <===ANSWER 3x^2 + 11x - 4 = 0 (multiply 3 and -4, which is -12; find two integers that multiply to -12 but add to 11, which are -1 and 12) 3x^2 + 12x - x - 4 = 0 (group into pairs) (3x^2 + 12x) + (-x - 4) = 0 (factor each group) 3x(x + 4) - 1(x + 4) = 0 (make (3x - 1) a factor and (x + 4) a factor) (3x - 1)(x + 4) = 0 (zero-product property) 3x - 1 = 0 or x + 4 = 0 3x = 1 or x = -4 x = 1/3 or x = -4 <===ANSWER Factoring by this method is called the AC method: http://www.tcc.edu/vml/Mth03/Trinom/tri7... It works on factoring any quadratic (if it is factorable to begin with, of course).

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