7 question solved please
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Just by applying the algebric identity=a2-b2=(a+b)(a-b) ..we have to factorise it ......Hope it helps..
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x²-3 = 0
where a= 1, b= 0 ( as there is no coefficient of x) and c= -3
(x+✓3)(x-✓3) = 0 ( as (a+b)(a-b) = a²-b²)
if x+✓3= 0 and x-√3 = 0, x= √3, -✓3
two zeroes are ✓3 and -✓3
now we know,
sum of the roots= (-b/a)
so putting values we get,
-✓3+✓3 = (-0/1)
0= 0
also, product of roots = (c/a)
putting the values we get,
(✓3)(-✓3) = (-3/1)
-3 = -3
hence verified
hope this helps you
where a= 1, b= 0 ( as there is no coefficient of x) and c= -3
(x+✓3)(x-✓3) = 0 ( as (a+b)(a-b) = a²-b²)
if x+✓3= 0 and x-√3 = 0, x= √3, -✓3
two zeroes are ✓3 and -✓3
now we know,
sum of the roots= (-b/a)
so putting values we get,
-✓3+✓3 = (-0/1)
0= 0
also, product of roots = (c/a)
putting the values we get,
(✓3)(-✓3) = (-3/1)
-3 = -3
hence verified
hope this helps you
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