Chapter 4 - Quadratic Equations, RD Sharma Solutions - (Part-3) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Q. 5. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
2x² + x + 4 = 0
Ans.
We have been given that, 
2x² + x + 4 = 0
Now divide throughout by 2. We get,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘x’ on both the sides. We have,


Since RHS is a negative number, therefore the roots of the equation do not exist as the square of a number cannot be negative.

Therefore the roots of the equation do not exist.

Q.6. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
x² – 8x + 18 = 0
Ans.
Given: x² – 8x + 18 = 0
x²−8x+18=0
Adding and subtracting (1/2 x 8)², we get
⇒ x²−8x+18+4²−4²=0

⇒ x²−8x+16+18−16=0
⇒ (x−4)²+2=0
⇒ (x−4)²=−2
⇒ (x−4) = ±√-2
But, √-2 is not a real number.
Hence, the roots of the quadratic equation doesn't exist.

Q.7. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
x²+4√3x+3=0
Ans.
We have been given that,
x²+4√3x+3=0
Now divide throughout by 4. We get,
x²+√3x+3/4=0
Now take the constant term to the RHS and we get
x²+√3x= -3/4
Now add square of half of co-efficient of ‘x’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.
So, now take the square root on both the sides and we get

Now, we have the values of ‘x’ as

Also we have,

Therefore the roots of the equation are

Q.8. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
√2x²−3x−2√2=0
Ans.
We have been given that,
√2x²−3x−2√2=0
Now divide throughout by √2. we get

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘x’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘x’ as

Also we have,

Therefore the roots of the equation are 2√2 and - 1/√2.

Q.9. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
√3x²+10x+7√3 = 0
Ans.
We have been given that,
√3x²+10x+7√3 = 0
Now divide throughout by√3. We get, 

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘x’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.
So, now take the square root on both the sides and we get

Now, we have the values of ‘x’ as

Also we have,

Therefore the roots of the equation are -√3 and - 7/√3.

Q.10. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
x²−(√2+1)x+√2 = 0
Ans.
We have been given that,
x²−(√2+1)x+√2 = 0
Now take the constant term to the RHS and we get
x²−(√2+1)x = -√2
Now add square of half of co-efficient of ‘x’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.
So, now take the square root on both the sides and we get

Now, we have the values of ‘x’ as

Also we have,

Therefore the roots of the equation are √2 and 1.

Q.11. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
x² - 4ax + 4a² - b² = 0
Ans.
We have to find the roots of given quadratic equation by the method of completing the square. We have,
x² - 4ax + 4a² - b² = 0
Now shift the constant to the right hand side,
x² - 4ax = b² - 4a²

Now add square of half of coefficient of x on both the sides,
x² - 2(2a)x + (2a)² = b² - 4a² + (2a)²
We can now write it in the form of perfect square as,
(x - 2a)² = b²
Taking square root on both sides,
(x - 2a) = √b²
So the required solution of x,
x = 2a ± b
= 2a+b, 2a - b

Page No 4.32

Q.1. Write the discriminant of the following quadratic equations:
(i) 2x² − 5x + 3 = 0
(ii) x² + 2x + 4 = 0
(iii) (x − 1) (2x − 1) = 0
(iv) x² − 2x + k = 0, k ∈ R
(v) √3x² + 2√2x - 2√3 = 0
(vi) x² − x + 1 = 0
(vii) (x + 5)² = 2(5x – 3)
Ans.
We have to find the discriminant of the following quadratic equations
(i) We have been given,
⇒ 2x² − 5x + 3 = 0
Now we also know that for an equation ax²+bx+c = 0 the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have  a = 2, b = -5 and c = 3.
Therefore, the discriminant is given as,
D = (-5)² - 4(2)(3)
= 25 - 24
= 1
Therefore, the discriminant of the equation is 1.

(ii) We have been given,
⇒ x² + 2x + 4 = 0
Now we also know that for an equation ax²+bx+c = 0 the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have  a = 1, b = 2 and c = 4.
Therefore, the discriminant is given as,
D = (2)² - 4(1)(4)
= 4 - 16
= -12
Therefore, the discriminant of the equation is -12

(iii) We have been given,
⇒ (x − 1) (2x − 1) = 0
Now, simplify the equation to be represented in the quadratic form, so we have
2x² - x - 2x + 1 = 0
2x² - 3x + 1 = 0
Now we also know that for an equation ax2 + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have,a = 2, b = -3 and c = 1.
Therefore, the discriminant is given as,
D = (-3)² - 4(2)(1)
= 9 - 8
= 1
Therefore, the discriminant of the equation is 1.

(iv) We have been given,
⇒ x² − 2x + k = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 1, b = -2 and c = k.
Therefore, the discriminant is given as,
D = (-2)² - 4(1)(k)
= 4 - 4k
Therefore, the discriminant of the equation is 4 - 4k.

(v) We have been given,
⇒ √3x² + 2√2x - 2√3 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = √3, b = 2√2 and c = -2√3.
Therefore, the discriminant is given as,
D=(2√2)²−4(√3)(−2√3)
= 8 + 24
= 32
Therefore, the discriminant of the equation is 32.

(vi) We have been given,
x² − x + 1 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 1, b = -1 and c = 1.
Therefore, the discriminant is given as,
D = (-1)² - 4(1)(1)
= 1 - 4
= -3
Therefore, the discriminant of the equation is -3.

(vii) We have been given,
(x + 5)² = 2(5x – 3)
(x+5)²=2(5x−3)
⇒x²+5²+2(x)(5)=10x−6
⇒x²+25+10x=10x−6
⇒x²+25+10x−10x+6=0
⇒x²+31=0
Discriminant
= b²−4ac                      
= 0²−4(1)(31)  
=−124
Hence, the discriminant is –124.

Q.2. In the following, determine whether the given quadratic equations have real roots and if so, find the roots:
(i) 16x² = 24x + 1
(ii) x² + x + 2 = 0
(iii) √3x²+10x−8√3=0
(iv) 3x² − 2x + 2 = 0
(v) 2x²−2√6x+3=0
(vi) 3a²x²+8abx+4b² = 0, a ≠ 0
(vii) 3x²+2√5x−5=0
(viii) x² − 2x + 1 = 0
(ix) 2x²+5√3x+6=0
(x) √2x²+7x+5√2=0
(xi) 2x²−2√2x+1=0
(xii) 3x² − 5x + 2 = 0
Ans.
In the following parts we have to find the real roots of the equations
(i) We have been given,
⇒ 16x² = 24x + 1
16x² - 24x - 1 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 16, b = -24 and c = -1.
Therefore, the discriminant is given as,
D = (-24)² - 4(16)(-1)
= 576 + 64
= 640
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are

(ii) We have been given,
⇒ x² + x + 2 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 1, b = 1 and c = 2.
Therefore, the discriminant is given as,
D = (1)² - 4(1)(2)
= 1 - 8
= -7
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation does not satisfies this condition, hence it does not have real roots.

(iii) We have been given,
⇒ √3x²+10x-8√3 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a =√3, b = 10 and c = -8√3.
Therefore, the discriminant is given as,
D=(10)²-4(√3)(-8√3)
= 100 + 96
= 196
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are 2/√3 and -4√3.

(iv) We have been given, 
⇒ 3x² - 2x + 2 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 3, b = -2 and c = 2.
Therefore, the discriminant is given as,
D = (-2)² - 4(3)(2)
= 4 - 24
= -20
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation does not satisfies this condition, hence it does not have real roots.

(v) We have been given,
⇒ 2x²- 2√6x + 3 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 2, b=-2√6 and c = 3.
Therefore, the discriminant is given as,
D=(-2√6)²-4(2)(3)
= 24 - 24
= 0
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real and equal roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Therefore, the roots of the equation are real and equal and its value is .

(vi) We have been given, 
⇒ 3a²x² + 8abx + 4b² = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 3a², b = 8ab and c = 4b².
Therefore, the discriminant is given as,
D = (8ab)² - 4(3a²)(4b²)
= 64a²b² - 48a²b²
= 16a²b²
Since, in order for a quadratic equation to have real roots, D ≥ 0. Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are

(vii) We have been given, 
⇒ 3x²+2√5x−5=0

Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 3, b=2√5 and c = -5.
Therefore, the discriminant is given as,
D=(2√5)²- 4(3)(-5)
= 20 + 60
= 80
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are

(viii) We have been given,
⇒ x² - 2x + 1 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 1, b = -2 and c = 1.
Therefore, the discriminant is given as,
D = (-2)² - 4(1)(1)
= 4 - 4
= 0
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real and equal roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

= 2/2
= 1
Therefore, the roots of the equation are real and equal and its value is 1.

(ix) We have been given, 
⇒ 2x²+5√3x+6=0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 2, b=5√3 and c = 6.
Therefore, the discriminant is given as,
D=(5√3)²-4(2)(6)
= 75 - 48
= 27
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are (-√3)/2 and -2√3.

(x) We have been given, 
⇒ √2x²+7x+5√2=0

Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b2 - 4ac
Now, according to the equation given to us, we have, a=√2, b = 7 and c=5√2.
Therefore, the discriminant is given as,

D=(7)²-4(√2)(5√2)
= 49 - 0
= 9
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

=-√2
Also,

Therefore, the roots of the equation are -5/√2 and -√2.

(xi) We have been given, 
⇒ 2x²−2√2x+1=0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 2, b=-2√2 and c = 1.
Therefore, the discriminant is given as,

D=(-2√2)²-4(2)(1)

= 8 - 8
= 0
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real and equal roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Therefore, the roots of the equation are real and equal and its value is 1/√2.

(xii) We have been given, 
⇒ 3x² − 5x + 2 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 3, b = -5 and c = 2.
Therefore, the discriminant is given as,

D = (-5)² - 4(3)(2)
= 25 - 24

= 1
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

x=(5+1)/6

=6/6
= 1
Also,
x=(5-1)/6
=4/6
=2/3
Therefore, the roots of the equation are 2/3 and 1.