Question

Question 2 Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Class 10 - Math - Quadratic Equations Page 87

Answer 1

Quadratic equation:

An equation of the form ax2+bx +c = 0 is called a quadratic equation in one variable, where a, b, c are real numbers and a ≠ 0.

Quadratic Formula:

This method is also called as Sridharacharya's rule.


x= –b±√b2 – 4ac/2a

where b2 - 4ac is called the discriminant of the quadratic equation and it is denoted by 'D'.

D= b2 - 4ac

x= -b±√D/2a


Nature of the roots


If D = 0 roots are real and equal , D > 0 roots are real and unequal, D < 0 roots are imaginary.

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Solution:

(i) 2x2 – 7x + 3 = 0

On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -7 and c = 3

By using quadratic formula, we get
x = –b±√b2 – 4ac/2a

⇒ x = 7±√49 – 24/4
⇒ x = 7±√25/4
⇒ x = 7±5/4
⇒ x = 7+5/4 or x = 7-5/4
⇒ x = 12/4 or 2/4

∴  x = 3 or 1/2

(ii) 2x2 + x – 4 = 0

On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = -4

By using quadratic formula, we get
x = –b±√b2 – 4ac/2a

⇒x = -1±√1+32/4
⇒x = -1±√33/4

∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4x2 + 4√3x + 3 = 0

On comparing this equation with ax2 + bx + c = 0, we get
a = 4, b = 4√3 and c = 3

By using quadratic formula, we get
x = –b±√b2 – 4ac/2a

⇒ x = -4√3±√48-48/8
⇒ x = -4√3±0/8

∴ x = √3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0

On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = 4

By using quadratic formula, we get
x = –b±√b2 – 4ac/2a

⇒ x = -1±√1-32/4

⇒ x = -1±√-31/4

The square of a number can never be negative.
∴there is no real solution of this equation.

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Hope this will help you.......

Answer 2

(i) 2x2 – 7x + 3 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -7 and c = 3
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = 7±√49 - 24/4
⇒ x = 7±√25/4
⇒ x = 7±5/4
⇒ x = 7+5/4 or x = 7-5/4
⇒ x = 12/4 or 2/4
∴  x = 3 or 1/2

(ii) 2x2 + x - 4 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = -4
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒x = -1±√1+32/4
⇒x = -1±√33/4
∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4x2 + 4√3x + 3 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 4, b = 4√3 and c = 3
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = -4√3±√48-48/8
⇒ x = -4√3±0/8
∴ x = -√3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 1 and c = 4
By using quadratic formula, we get
x = -b±√b2 - 4ac/2a
⇒ x = -1±√1-32/4
⇒ x = -1±√-31/4
The square of a number can never be negative.
∴There is no real solution of this equation.