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Qudratic equations..x^2 - 4x = 12
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imaginary roots.no real roots
\[x^2-4x-12=0\] \[(x-6)(x+2)=0\] so \[x=6\ or \ x=-2\]
Ok that what I got but the answer sheet is telling me 2,6,12,16
solve it using completing the square method add square of half of the coefficient of x i.e. 4/2 = 2 to both sides x² - 4x + 2² = 12 + 2² LHS takes the form a² - 2ab + b² = (a - b)² (x - 2)² = 16 taking root of both sides x-2 = ± root(16) x = 2 ± 4 x = 2+4 =6 and x = 2-4 = -2
x = 6, x = -2
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The answer sheet is wrong.
6
or -2
u can substitute the values into the equation and calculate... you will that -2 and 6 r the right answers..
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