◆Check whether x²+2√x-3=0 is a quadratic equation or not ?
◆Find the roots of (x+7) (x+4)=0
◆Solve 5x²-17x+12=0 by factorisation method.
◆Solve a²b²x²+b²x-a²x-1 =0
rohitkumargupta:
check your first question
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(1):
If it is a quadratic equation, it must be in the form of (ax²+bx+c=0) and a≠0
As we can observe that
ax² = x² where a = 1
2√x = bx where b = 2
-3 = c where c = -3
Hence, this is a quadrilateral equation.
(2):
(x + 7)(x + 4) = 0
By Zero Product Rule, Anything multiplied by 0 is equal to 0, So (x + 7) can be 0 or (x + 4) can be 0
=> (x + 7 = 0) or (x + 4 = 0)
=> (x = -7) or (x = -4)
(3):
5x² - 17x + 12 = 0
5x² - (12 + 5)x + 12 = 0
5x² - 12x - 5x + 12 =0
x(5x - 12) - 1(5x - 12) = 0
(x - 1)(5x - 12) = 0
(x = 1). Or. ( x = 12/5)
(4):
If it is a quadratic equation, it must be in the form of (ax²+bx+c=0) and a≠0
As we can observe that
ax² = x² where a = 1
2√x = bx where b = 2
-3 = c where c = -3
Hence, this is a quadrilateral equation.
(2):
(x + 7)(x + 4) = 0
By Zero Product Rule, Anything multiplied by 0 is equal to 0, So (x + 7) can be 0 or (x + 4) can be 0
=> (x + 7 = 0) or (x + 4 = 0)
=> (x = -7) or (x = -4)
(3):
5x² - 17x + 12 = 0
5x² - (12 + 5)x + 12 = 0
5x² - 12x - 5x + 12 =0
x(5x - 12) - 1(5x - 12) = 0
(x - 1)(5x - 12) = 0
(x = 1). Or. ( x = 12/5)
(4):
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