In the following, determine whether the given quadratic equations have real roots and if so, find the roots:
(iv)3x² − 2x + 2 = 0
(v)2x²-2√6x+3=0
(vi)3a²x²+8abx+4b²=0, a≠0
Answers
SOLUTION :
(iv) Given : 3x² − 2x + 2 = 0
On comparing the given equation with, ax² + bx + c = 0
Here, a = 3 , b = -2 , c = 2
Discriminant , D = b² - 4ac
D = (-2)² - 4(3)(2)
D = 4 - 24
D = - 20
Since, D < 0 so given Quadratic equation has no real roots .
(v) Given : 2x² - 2√6x + 3 = 0
On comparing the given equation with, ax² + bx + c = 0
Here, a = 2 , b = -2√6 , c = 3
Discriminant , D = b² - 4ac
D = (-2√6)² - 4 × 2 × 3
D = 24 - 24
D = 0
Since, D = 0, so given Quadratic equation has two equal and real roots which are given by
x = [- b ± √D]/2a
x= −(-2√6) ±√0 / 2 × 2
x =( 2√6 ± 0)/4
x = 2√6/4
x = √6/2
x = (√3 √2)/ (√2 √2) = √3/2
x = √(3/2)
Hence, the Roots are √(3/2) and √(3/2) .
(vi) Given : 3a²x² + 8abx + 4b² = 0, a ≠ 0
On comparing the given equation with, ax² + bx + c = 0
Here, a = 3a² , b = 8ab , c = 4b²
Discriminant , D = b² - 4ac
D = (8ab)² - 4 × 3a² × 4b²
D = 64a²b² - 48a²b²
D = 16a²b²
Since, D ≥ , 0 so given Quadratic equation has distinct real roots which are given by
x = [- b ± √D]/2a
x= [ −(8ab) ±√16a²b²] / 2 ×3a²
x = (- 8ab ± 4ab) /6a²
x = 2a(- 4b ± 2b/ 6a²
x = (- 4b ± 2b)/ 3a
x = (- 4b + 2b)/ 3a or x = (- 4b - 2b)/ 3a
x = -2b/3a or x = - 6b/3a
x = -2b/3a or x = - 2b/a
Hence, the Roots are -2b/3a and - 2b/a.
HOPE THIS ANSWER WILL HELP YOU...
Solution :
_______________________
Nature of roots of a
quadratic equation
ax²+bx+c=0, a≠0 is
Discreminant (D)
= b²-4ac
If i ) D > 0
Roots are real and distinct.
ii ) D = 0
Roots are real and equal.
iii ) D < 0
Roots are not real.
_____________________
(iv)3x² − 2x + 2 = 0
Compare this with
ax²+bx+c =0 , we get
a = 3 , b = -2 , c = 2
D = b² - 4ac
= (-2)² - 4×3×2
= 4 - 24
= -20
D < 0
Therefore ,
Roots are not real.
(v)2x²-2√6x+3=0
a = 2 , b = -2√6 , c = 3
D = (-2√6)² - 4×2×3
= 24 - 24
D = 0
Therefore ,
Roots are real and equal.
(vi)3a²x²+8abx+4b²=0, a≠0
D = (8ab)² - 4×3a²×4b²
= 64a²b² - 48a²b²
= 16a²b²
D = (4ab )²
D > 0
Therefore ,
Roots are real and distinct.
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