Question

Find the nature of the roots of the following quadratic equations. If real roots exist, find them:
(i) 2x² – 3x + 5 = 0 (ii) 3x² − 4√3x + 4 = 0 (iii) 2x² – 6x + 3 = 0

Answer 1

For a quadratic equation ax² + bx + c =0, the term b² - 4ac is called discriminant (D) of the quadratic equation because it  determines  whether the quadratic equation has real roots or not ( nature of roots).

D=  b² - 4ac

So, a quadratic equation ax² + bx + c =0, has

i) Two distinct real roots, if b² - 4ac >0 , then x= -b/2a + √D/2a  &x= -b/2a - √D/2a

ii) Two equal real roots, if b² - 4ac = 0 , then x= -b/2a or -b/2a

iii) No real roots, if b² - 4ac <0

SOLUTION ::

i) x² – 3x + 5 = 0

On Comparing it with ax² + bx + c = 0, we get

Here, a = 2, b = -3 and c = 5

Discriminant (D) = b² – 4ac

D = (- 3)² - 4 (2) (5)  

D = 9 – 40

D= – 31<0

As b² – 4ac < 0,

Hence, no real root is possible .

(ii) 3x² – 4√3x + 4 = 0

On Comparing it with ax² + bx + c = 0, we get

Here,a = 3, b = -4√3 and c = 4

Discriminant(D) = b² – 4ac

D= (-4√3)2 – 4(3)(4)

D= 48 – 48 = 0

As b² – 4ac = 0,

Hence,  real roots exist & they are equal to each other.

roots will be –b/2a and –b/2a.

-b/2a = -(-4√3)/2×3  

= 4√3/6  

= 2√3/3  

multiplying the numerator & denominator by √3

= (2√3) (√3) / (3)(√3)  

= 2 ×3 / 3 ×√3  

= 2/√3

Hence , the equal roots are 2/√3 and 2/√3.

(iii) 2x² – 6x + 3 = 0

On Comparing this equation with ax² + bx + c = 0, we get

Here,a = 2, b = -6, c = 3

Discriminant (D)= b² – 4ac

D= (-6)2 – 4 (2) (3)

D= 36 – 24 = 12

As b2 – 4ac > 0,

Hence, two distinct real roots exist for this equation

x= -b/2a + √D/2a  &     x= -b/2a - √D/2a

x= (6+√12) / 2×2

x= 6+√4×3 /4  

x= 6 + 2√3 /4  

x= 2( 3 + √3) 4  

x = 3 + √3 /2

x= (6-√12) / 2×2

x= 6-√4×3 /4  

x= 6 - 2√3 /4  

x= 2(3 - √3)/ 4  

x= 3 - √3 /2

Hence the real roots are 3 + √3 /2   & 3 -√3 /2

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Dear Student,

Solution:

To find nature of roots,we first calculate Determinant D.

D =
${b}^{2} - 4ac$
here a,b and c are the coefficient of x^2,x and constant term respectively

D = 0; roots are real and equal

D >0; roots are real and distinct

D <0; No real roots exists

1) 2x² – 3x + 5 = 0

compare it with
$a {x}^{2} + bx + c = 0$
here a= 2 ,b = -3 and c = 5
D

$= {( - 3)}^{2} - 4(2)(5) \\ = 9 - 40 \\ = - 31$
since D<0, thus equation 2x² – 3x + 5 = 0 dies not have real roots.

2) 3x² − 4√3x + 4 = 0

compare it with standard equation again you

found that here a= 3 ,b = -4√3 and c = 4

D
$= ( { - 4 \sqrt{3} )}^{2} - 4(3)(4) \\ \\ = 48 - 48 \\ \\ = 0$
Here D = 0 ,since equation has real and equal roots.

these roots are same and given as
$x = \frac{ - b}{2a}$
x1,2 = 4√3/6 = 2√3/3

roots are (2√3)/3

3)2x² – 6x + 3 = 0

a = 2

b = -6

c = 3
D
$= ( { - 6)}^{2} - 4(2)(3) \\ = 36 - 24 \\ = 12$

since D > 0, roots are real and distinct

$x1 = \frac{ 6+ \sqrt{ 12 } }{4} \\ \\ x1 = \frac{3 + \sqrt{3} }{2} \\ \\ x2 = \frac{3 - \sqrt{3} }{2}$
Hope it helps you.