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Find the discriminant of the following quadratic equations:
2x^{2} - 5x + 3 = 0
The given equation is in the form of ax^{2} + bx + c = 0
Here, a = 2, b = -5 and c = 3
The discriminant, D = b^{2} - 4ac
D = (-5)^{2} – 4 x 2 x 3
D = 25 – 24 = 1
Therefore, the discriminant of the following quadratic equation is 1.
Find the discriminant of the quadratic equations.
x^{2} + 2x + 4 = 0
Here, a = 1, b = 2 and c = 4
The discriminant is:-
D = (2)^{2} - 4 x 1 x 4
D = 4 – 16 = - 12
The discriminant of the following quadratic equation is = - 12.
(x -1) (2x -1) = 0
The provided equation is (x -1) (2x -1) = 0
By solving it, we get 2x^{2} - 3x + 1 = 0
Now this equation is in the form of ax^{2} + bx + c = 0
Here, a = 2, b = -3, c = 1
D = (-3)^{2} - 4 x 2 x 1
D = 9 - 8 = 1
The discriminant of the following quadratic equation is = 1.
x^{2 }- 2x + k = 0
Here, a = 1, b = -2, and c = k
D = b^{2} - 4ac
D = (-2)^{2} - 4(1)(k)
= 4 - 4k
Therefore, the discriminant, D of the equation is (4 - 4k)
The discriminant is, D = b^{2} - 4ac
D = 8 + 24 = 32
The discriminant, D of the following equation is 32.
x^{2} - x + 1 = 0
Here, a = 1, b = -1 and c = 1
The discriminant is D = b^{2 }- 4ac
(-1)^{2} - 4 × 1 × 1
1 - 4 = - 3
Therefore, the discriminant D of the following equation is -3.
x^{2} + x + 2 = 0
The given equation can be written in the form of, ax^{2 }+ bx + c = 0
The discriminant is given by the following equation, D = b^{2} - 4ac
Here, a = 1, b = 1 and c = 2.
Therefore, the discriminant is given as,
D = (1)^{2} - 4(1)(2)
= 1 - 8
= - 7
For a quadratic equation to have real roots, D ≥ 0.
Here we find that the equation does not satisfy this condition, hence it does not have real roots.
16x^{2} = 24x + 1
16x^{2} - 24x - 1 = 0
The discriminant is given by the following equation, D = b^{2 }- 4ac
Here, a = 16, b = -24 and c = - 1
D = (-24)^{2} - 4(16)(-1)
= 576 + 64
= 640
Here it can be seen that the equation satisfies this condition, hence it has real roots.
The roots of an equation can be found out by using,
Therefore, the roots of the following equation are as follows,
The values of x for both the cases will be:
For a quadratic equation to have real roots, D ≥ 0
Therefore, the roots of the equation are given as follows,
3x^{2} - 2x + 2 = 0
Here, a = 3, b = -2 and c = 2.
D = (-2)^{2} - 4(3)(2)
= 4 - 24 = – 20
Here it can be seen that the equation does not satisfy this condition, hence it has no real roots.
= 24 - 24 = 0
3a^{2}x^{2 }+ 8abx + 4b^{2} = 0
Here, a = 3a^{2}, b = 8ab and c = 4b^{2}
D = (8ab)^{2} - 4(3a^{2})(4b^{2})
= 64a^{2}b^{2} - 48a^{2}b^{2 }= 16a^{2}b^{2}
= 20 + 60
= 80
x^{2} - 2x + 1 = 0
Here, a =1, b = - 2 and c = 1
D = (-2)^{2} - 4(1)(1)
= 4 - 4
= 0
x = 2/2
x = 1
Therefore, the equation real roots and its value is 1
= 75 - 48
= 27
The roots of an equation can be found out by using,\
Therefore, the roots of the equation are given as follows
The values of x for both the cases will be
D = 49 - 40
D = 9
= 8 - 8
3x^{2} - 5x + 2 = 0
Here, a = 3, b = -5 and c = 2.
D = (-5)^{2} - 4(3)(2)
= 25 - 24
= 1
And,
x = 2/3
Solve for x:
The above equation can be solved as follows:
6x^{2} - 30x + 30 = 10x^{2} - 60x +80
4x^{2} - 30x + 50 = 0
2x^{2} - 15x + 25 = 0
The above equation is in the form of ax^{2} + bx + c = 0
The discriminant is given by the equation, D = b^{2 }- 4ac
Here, a = 2, b = -15, c = 25
D = (-15)^{2} - 4(2)(25)
= 225 - 200
= 25
x = 5
Also,
x = 5/2
x^{2} + 1 = 3x
x^{2} - 3x + 1 = 0
Here, a = 1, b = - 3, c = 1
D = (-3)^{2 }- 4(1)(1)
D = 9 - 4
D = 5
(16 - x)(x + 1) = 15x
16x + 16 - x^{2} - x = 15x
15x + 16 - x^{2} - 15x = 0
16 - x^{2} = 0
X^{2} - 16 = 0
Here, a = 1, b = 0, c = -16
D = (0)^{2} - 4(1)(-16)
D = 64
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Chapter 8: Quadratic Equations Exercise –...