Calculate the value of the discriminant of x^2+2x+1=0?
回答 (10)
2008-07-29 2:41 pm
✔ 最佳答案
In the equation ax^2+bx+c, the discriminant is b^2-4ac.Therefore, (2)^2-4x1x1
4-4
0
The discriminant is 0. This means the equation has one real double-root.
2008-07-29 2:41 pm
don't you know it is a perfect square?
its discriminant is 0.
its discriminant is 0.
2008-07-29 2:41 pm
The discriminant is part of the quadratic formula.
(-b +- sqrt(b^2 - 4ac)) / 2a
The discriminant is the part under the radical, or
b^2 - 4ac
a = 1
b = 2
c = 1
b^2 - 4ac =
2^2 - 4*1*1 = 4 - 4 = 0
By the way this equation only has one solution, x = 1
(-b +- sqrt(b^2 - 4ac)) / 2a
The discriminant is the part under the radical, or
b^2 - 4ac
a = 1
b = 2
c = 1
b^2 - 4ac =
2^2 - 4*1*1 = 4 - 4 = 0
By the way this equation only has one solution, x = 1
2008-07-29 2:40 pm
b^2 - 4*a*c
a=1b=2 c=1
2^2 - 4*1*1 = 4- 4=0
which is a square number. so this question was factorable
a=1b=2 c=1
2^2 - 4*1*1 = 4- 4=0
which is a square number. so this question was factorable
2016-11-14 5:38 pm
[3] the standard form of a quadratic equation is ax^2+bx+c=0,the place a is the coefficient of x^2,b is the coefficient of x and c is a persevering with. in accordance to quadratic formula,the fee/s of x could be found out from the formula x={-b+-sqrt(b^2-4ac)}/2a Now,the area b^2-4ac of the formula is primary as discriminant as from it ,we are able to appreciate the character of the roots of x subsequently,via comparing with usual quadratic equation ax^2+bx+c=0.we get a=a million,b= -2 and c=a million The discriminant =b^2-4ac =(-2)^2-4*a million*a million =4-4 =0 be conscious-through fact the discriminant is 0,we are able to assert that there will be purely one root of the equation
2008-07-29 2:40 pm
b^2-4ac=2^2 -4(1)(1)=0
one real root
one real root
2008-07-29 3:00 pm
x^2 + 2x + 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a
a = 1
b = 2
c = 1
x = [-2 ±√(4 - 4)]/2
x = [-2 ±√0]/2
x = -2/2
x = -1
∴ x = -1
x = [-b ±√(b^2 - 4ac)]/2a
a = 1
b = 2
c = 1
x = [-2 ±√(4 - 4)]/2
x = [-2 ±√0]/2
x = -2/2
x = -1
∴ x = -1
2008-07-29 2:57 pm
1) By factorization method:
x^2 + x + x + 1 = 0
x (x+1) +1 (x+1) = 0
(x+1) (x+1)=0
so,
x + 1 = 0
x = -1
2) By completing square method:
x^2 + 2x + 1 = 0
x^2 + 2x = -1
adding 1^2 on both sides
x^2 + 2x + 1^2 = -1 + 1^2
(x+1)^2 = -1 + 1
(x+1)^2 = 0
√(x+1)^2 = √0
x + 1 = 0
x = -1
x^2 + x + x + 1 = 0
x (x+1) +1 (x+1) = 0
(x+1) (x+1)=0
so,
x + 1 = 0
x = -1
2) By completing square method:
x^2 + 2x + 1 = 0
x^2 + 2x = -1
adding 1^2 on both sides
x^2 + 2x + 1^2 = -1 + 1^2
(x+1)^2 = -1 + 1
(x+1)^2 = 0
√(x+1)^2 = √0
x + 1 = 0
x = -1
2008-07-29 2:46 pm
discriminant= (coefficient of x term)^2 - 4(coefficient of x^2 term)(constant)
so in this case,
discriminant= (2)^2 - 4(1)(1)
=4-4
=0
(as a note, this means that the roots of the equation - the solutions for x - are real and equal. In this case both x= -1)
so in this case,
discriminant= (2)^2 - 4(1)(1)
=4-4
=0
(as a note, this means that the roots of the equation - the solutions for x - are real and equal. In this case both x= -1)
2008-07-29 2:41 pm
(x + 1) (x + 1) = 0
So there is 1 solution, x = -1
So there is 1 solution, x = -1
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