Quadratic Equation
Mathematics Class Tenth
NCERT Exercise 4.3
Question (1) Find the roots of the following quadratic equations, if they exist, by the method of completing the square
Question (1) (i) 2 x2 – 7 x + 3 = 0
Solution
Given, 2 x2 – 7 x + 3 = 0
To check whether the roots of the given quadratic equation exist or not, we need to calculate the b2 – 4ac
In the given equation, a = 2, b = 7 and c = 3
∴ b2 – 4ac = 72 – 4 × 2 × 3
= 49 – 24 = 25
Since, b2 – 4ac > 0 thus roots of the given equation exist
Now given equation is
2x2 – 7x + 3 = 0
In order to complete the square, after dividing the equation by 2 on both sides, we get
x2 – 7/2x + 3/2 = 0
⇒ x2 – 2 (7/2) x + 3/2 = 0
⇒ x2 – 2(7/2) x = – 3/2
In order to complete square, after adding (7/4)2 both sides, we get
x2 – 2(7/2) x + (7/4)2 = (7/4)2 – 3/2
⇒ (x – 7/4)2 = 49/16 – 3/2
⇒ (x – 7/4) 2 = 49 – 24/16
⇒ (x – 7/4)2 = 25/16
⇒ x – 7/4 = 25/16
⇒ x – 7/4 = ± 5/4
⇒ x = 5/4 + 7/4 or x = – 5/4 + 7/4
⇒ x = 12/4 and x = 2/4
⇒ x = 3 or x = 1/2
Roots of the given quadratic equation are 3 and 1/2
Question (1) (ii) 2 x2 + x – 4 = 0
Solution:
Given, 2 x2 + x – 4 = 0
To check whether the roots of the given quadratic equation exist or not, we need to calculate the D which is = b 2 – 4 a c
In the given equation, a = 2, b = 1 and c = – 4
∴ D = b2 – 4 a c
⇒ D =12 – 4 × 2 ×( – 4)
⇒ D = 1 + 32 = 33
Since, D > 0 thus two roots of the given equation exist
Now, given equation is 2 x2 + x – 4 = 0
In order to complete the square, after taking 2 as common, we get
⇒ 2(x2 + x/2 – 4/2) = 0
⇒ x2 + 1/2 x – 2 = 0
⇒ x2 + 2(1/4)x – 2 = 0
⇒ x2 + 2 (1/4) x = 2
In order to complete square, after adding (1/4)2 both sides, we get
⇒ x2 + 2 (1/4) x + (1/4)2 = 2 +(1/4 )2
⇒ (x + 1/4)2 = 2 + 1/16
⇒ (x + 1/4 )2 = 2 + 1/16
⇒ (x + 1/4 )2 = 32 + 1/16
⇒ x + 1/4 = 33/16
⇒ x + 1/4 = ± 33/4
⇒ x = ± 33/4 – 1/4
⇒ x = ± 33 – 1/4
⇒ x = ± 33/4 – 1/4
⇒ x = – 1 ± 33/4
Thus, roots are – 1 + 33/4 and – 1 – 33/4 Answer
Question (1) (iii) 4 x2 + 4√3 x + 3 = 0
Solution:
Given, 4 x2 + 4 √3 x + 3 = 0
To check whether the roots of the given quadratic equation exist or not, we need to calculate the D which is = b2 – 4 a c
In the given equation, a = 4, b = 4 √ 3 and c = 3
∴ D =(4 √3)2 – 4 × 4 × 3
⇒ D = 48 – 48 = 0
Since, D = 0, thus only one root is possible for the given equation.
Now, 4 x2 + 4 √3 x + 3 = 0
After taking 4 as common
⇒ 4(x2 + 4√ 3/4 x + 3/4) = 0
⇒ x2 + √ 3 x + 3/4 = 0
In order to complete the square after multiplying and dividing the middle term by 2
⇒ x2 + 2 (√3/2) x + 3/4 = 0
⇒ x2 + 2 (√ 3 /2) x = – 3/4
In order to complete square, after adding (√ 3 /2)2 both sides, we get
x2 + 2 (√ 3 /2) x + (√ 3 /2)2 = – 3/4 + (√ 3 /2)2
⇒ (x + √ 3 /2)2 = – 3/4 + 3/4
⇒ (x + √3/2)2 = 0
⇒ x + √3/2 = 0
⇒ x = – √3/2
Thus, root is – √3/2 Answer
Question(1)(iv) 2x2 + x + 4 = 0
Solution:
To check whether the roots of the given quadratic equation exist or not, we need to calculate the D which is = b2 – 4ac
In the given equation, a = 2, b = 1 and c = 4
∴ D = 12 – 4 × 2 × 4
⇒ D = 1 – 32 = – 31
Since, D < 0, thus, no real root will exist for the given quadratic equation Answer
Question (2) Find the roots of the quadratic equations given in Q. 1 above by applying the quadratic formula
Question (2)(i) 2x2 – 7x + 3 = 0
Solution:
Given, 2x2 – 7x + 3 = 0
Here, a = 2, b = – 7 and c = 3
We know that for a quadratic equation ax2 + b x + c = 0
roots are given by – b ±√b2 – 4 a c/2 a
Thus, roots of the given quadratic equation
– (– 7) ±√(– 7)2 – 4 ×2 × 3/2 ×2
= 7 ±√49 – 24/4
= 7 ±√25/4
=7± 5/4
=(7 + 5)4 and 7 – 5/4
= 12/4 and 2/4
= 3 and 1/2
Thus, roots are 3 and 1/2Answer
Question(2)(ii) 2x2 + x – 4 = 0
Solution:
Given, 2x2 + x – 4 = 0
Here, a = 2, b = 1 and c =-4`
We know that for a quadratic equation ax2 + bx + c = 0
roots are given by – b ±√b2 – 4 a c/2 a or – b ±√(D)/2 a
Where, D = b2 – 4ac
After putting the values of a, b, and c, we get
D = 12 – 4 × 2 × (– 4)
⇒ D = 1 – 8 × (– 4)
⇒ D = 1 + 32 = 33
Thus, roots of the given equation
=– b ±√(D)/2 a
After putting the value of D, b, and a, we get
– 1 ±√33/2 × 2
= – 1 ±√33/4
Thus, Roots are – 1 + √33/4 and – 1 – √33/4 Answer
Question (2)(iii) 4x2 + 4 (3x) + 3 = 0
Solution:
Given, 4x2 + 4(3x) + 3 = 0
Here, a = 4, b = 4 √3, and c = 3
We know that for a quadratic equation ax2 + bx + c = 0
roots are given by – b ±√b2 – 4 a c/2 a or – b ±√(D)/2 a
Where, D = b2 – 4 a c
After putting the values of a, b, and c, we get
D =(4(3)2 – 4 × 4 × 3
⇒ D = 16 × 3 – 48
⇒ D = 48 – 48 = 0
Thus, root = – b ±√(D)/2 a
= – 4 ±√ 0/2 × 4
= – 43/8 = 3/2
Thus, root for given equation is sqrt3/2
Question(2) (iv) 2x2 + x + 4 = 0
Solution:
Given, 2x2 + x + 4 = 0
Here, a = 2, b = 1 and c = 4
We know that for a quadratic equation ax2 + bx + c = 0
roots are given by – b ±√b2 – 4 a c/2 a
or – b ±√(D)/2 a
Where, D = b2 – 4ac
After putting the values of a, b and c, we get
D = 12 – 4 × 2 × 4
⇒ D = 1 – 32 = – 31
Here, since D < 0, thus no root is possible for the given quadratic equation
Question (3) Find the roots of the following equations:
Question (3)(i) x – 1/x = 3, ≠ = 0
Solution:
Given, x – 1/x = 3
⇒ x – 1/x – 3 = 0
⇒ x2 – 1 – 3x/x = 0
⇒ x2 – 3x – 1 = 0
Here, a = 1, b = – 3 and c = – 1
We know that for a quadratic equation ax2 + bx + c = 0
roots are given by – b ±√b2 – 4 a c/2 a
or – b ±√(D)/2 a
Thus, by putting values of a, b and c
Roots = – (– 3) ±√(– 3)2 – 4 × 1 × (– 1)/2 × 1
= 3± 9 + 4/2
= 3 ± 13/2`
∴ 3 + 13/2 and 3 – 13/2 are the roots of the given quadratic equation
Question (3)(ii) 1/x + 4 – 1/x – 7 = 11/30, x ≠ = – 4, 7
Solution:
Given, 1/x + 4 – 1/x – 7 = 11/30
⇒ x – 7 – (x + 4)/(x + 4)(x – 7) = 11/30
⇒ x – 7 – x – 4/( x + 4)(x – 7) = 11/30
⇒ – 11/(x + 4)(x – 7) = 11/30
After cross multiplication
⇒ 11(x + 4)(x – 7)= – 11 × 30
⇒ 11(x2 – 7x + 4x – 28) = – 330
⇒ 11(x2 – 3x – 28) = – 330
⇒ 11x2 – 33x – 308 = – 330
⇒ 11x2 – 33x – 308 + 330 = 0
⇒ 11x2 – 33x + 22 = 0
By taking 11 as common
– 11(x2 – 3x + 2) = 0
⇒ x2 – 3x + 2 = 0
By expanding middle term
⇒ x2 – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2)= 0
⇒ (x – 1)(x – 2) = 0
If x – 1 = 0
∴ x = 1
And if x – 2 = 0
∴ x = 2
Thus, roots of the given quadratic equation is equal to 1 and 2
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