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9966 [12]
3 years ago
7

MATH URGENT

Mathematics
1 answer:
maks197457 [2]3 years ago
7 0

{x}^{2}   + 4x - 12 = 0 \\  {x}^{2}  + 4x + 4 - 4 - 12  = 0\\   {(x + 2)}^{2}  - 16 = 0 \\ x =   -   +  \sqrt{16}  - 2 \\ x =   + \:  or - 4 - 2 \\ x =  - 6 \:  \: x = 2
- subtract the 12 from both sides so that it becomes the last constant term in the quadratic equation which should now equal 0.
- take the 4x
- half the coefficient of 4 (2)
- square it (4)
- add it to the equation (+4)
- subtract it from the equation (-4)
- factorise the square (x+2)^2 expands to (x^2 + 4x + 2) as {a+b}^2={a^2 + ab + ba + b^2}
- now the equation is in turning point form.
- to find x, add 16 and square root 16 and (x+2)
- subtract 2 from positive or negative 4 (as -4^2 and 4^2 both equal 16).
- This should give you two values for x, -6 and 2.


I really hope that this helped :)
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