Question -
Answer -
(i) 2x2 –3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solution:
(i) Given,
2x2 – 3x + 5 = 0
Comparing the equation with ax2 + bx + c =0, we get
a = 2, b =-3 and c = 5
We know, Discriminant = b2 –4ac
= (– 3)2 – 4 (2) (5) = 9 – 40
= – 31
As you can see, b2 – 4ac <0
Therefore, no real root is possible for thegiven equation, 2x2 – 3x + 5 = 0.
Solution
(ii) 3x2 – 4√3x +4 = 0
Comparing the equation with ax2 + bx + c =0, we get
a = 3, b = -4√3 and c =4
We know, Discriminant = b2 –4ac
= (-4√3)2 – 4(3)(4)
= 48 – 48 = 0
As b2 – 4ac =0,
Real roots exist for the given equation andthey are equal to each other.
Hence the roots will be –b/2a and –b/2a.
–b/2a = -(-4√3)/2×3 = 4√3/6= 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and2/√3.
Solution
(iii) 2x2 – 6x +3 = 0
Comparing the equation with ax2 + bx + c =0, we get
a = 2, b =-6, c = 3
As we know, Discriminant = b2 –4ac
= (-6)2 – 4 (2) (3)
= 36 – 24 = 12
As b2 – 4ac >0,
Therefore, there are distinct real roots existfor this equation, 2x2 – 6x + 3 = 0.
= (-(-6) ± √(-62-4(2)(3)) )/ 2(2)
= (6±2√3 )/4
= (3±√3)/2
Therefore the roots for the given equation are(3+√3)/2 and (3-√3)/2