Question

Question 1 Find the nature of the roots of the following quadratic equations. If the real roots exist, find them; (I) 2x^2 −3x + 5 = 0 (II) 3x^2 - 4.3^0.5x + 4 = 0 (III) 2x^2 − 6x + 3 = 0

Class 10 - Math - Quadratic Equations Page 88

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

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Solution:

i)

x² – 3x + 5 = 0

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

a = 2, b = -3 and c = 5

Discriminant (D) = b² – 4ac

⇒ ( – 3)2 – 4 (2) (5) = 9 – 40
⇒ – 31<0

As b2 – 4ac < 0,

Hence, no real root is possible .

(ii) 3x² – 4√3x + 4 = 0

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

a = 3, b = -4√3 and c = 4

Discriminant(D) = b² – 4ac

⇒ (-4√3)2 – 4(3)(4)

⇒ 48 – 48 = 0

As b² – 4ac = 0,
Hence,  real roots exist & they are equal to each other.

the 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

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

a = 2, b = -6, c = 3

Discriminant (D)= b² – 4ac

= (-6)2 – 4 (2) (3)
= 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= 6+√4×3 /4  = 6 + 2√3 /4  = 2( 3 + √3) 4  = 3 + √3 /2

x = 3 + √3 /2

x= (6-√12) / 2×2= 6-√4×3 /4  = 6 - 2√3 /4  = 2( 3 - √3) 4  = 3 - √3 /2

x= 3 - √3 /2

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

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Hope this will help you....

Answer 2

(i) Consider the equation
x2 - 3x + 5 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 - 4ac
= ( - 3)2 - 4 (2) (5) = 9 - 40
= - 31
As b2 - 4ac < 0,
Therefore, no real root is possible for the given equation.

(ii) 3x2 - 4√3x + 4 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant = b2 - 4ac
= (-4√3)2 - 4(3)(4)
= 48 - 48 = 0
As b2 - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And 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 and 2/√3.

(iii) 2x2 - 6x + 3 = 0

Comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac

= (-6)2 - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation:

x = -b ± b² - 4ac / 2a
= -(-6) ± √(-6)²- 4(2)(3) / 2 × 2

= 6 ± √12 / 4
= 6 ± 2√3/ 4
= 3 ± √3 / 2



Therefore, the roots are 3+√3/2 and 3-√3/2