Question

Find the nature of the roots of the given equation; if it is real or equal then find

roots of the equation, 4x2 – 3x + 1 = 0.​

Answer 1

Explanation :

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

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