find zeroes of x^2-18x+12.
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when finding the roots we take equation is equal to zero because when we fill the value of the zeroes then it equal to zero
x²-18x+12 = 0
it is in a form of ax²+bx+c = 0 so,
a = 1; b= -18; c = 12.
_______________________
x = -b±√(b²-4ac)/2a
x = -(-18)±√(-18)²-4(1)(12)/2(1)
x = 18±√324-48/2
x = 18±√276/2
x = 18±√69×4/2
x = 18±2√69/2
x = 2(9±√69)/2
x = 9±√69
x = 9+√69 . Or. x = 9-√69
__________________________
x²-18x+12 = 0
it is in a form of ax²+bx+c = 0 so,
a = 1; b= -18; c = 12.
_______________________
x = -b±√(b²-4ac)/2a
x = -(-18)±√(-18)²-4(1)(12)/2(1)
x = 18±√324-48/2
x = 18±√276/2
x = 18±√69×4/2
x = 18±2√69/2
x = 2(9±√69)/2
x = 9±√69
x = 9+√69 . Or. x = 9-√69
__________________________
krinajoshi04p5kzdu:
thx bro
Kindly correct the last 2nd step. 9±√69/4
but there is no option of edit
Given....
thank you so much by mistake it happen
Answer should be - 9+√69 and 9-√69. 18±2√69/4 should be 18±2√69/2.
please give last chance
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