6. Chapter 6. Quadratic Equations NIOS Class 10 Mathematics Textbook Solutions
Chapter 6. Quadratic Equations NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 6. QUADRATIC EQUATIONS
CHECK YOUR PROGRESS 6.1
1. Which of the following equations are quadratic equations ?
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Solution: (i) 
is not a quadratic equation as 
is not a quadratic polynomial.
(ii) 
is a quadratic equation as 
is a quadratic polynomial.
(iii) We have,
is a quadratic equation as 
is a quadratic Polynomial.
(iv) We have,
is a quadratic equation as 
is a quadratic polynomial .
(v) We have,
is a quadratic equation as 
is a quadratic polynomial.
CHECK YOUR PROGRESS 6.2
1. Which of the following quadratic equations are in standard form? Those, which are not in standard form, rewrite them in standard form:
(i) 
(ii) 
(iii) 
(iv) 
[Note: The standard form of the quadratic equation is 
]
Solution: (i)
is a standard form .
(ii) is in the standard form .
is a standard form.
(iii) is not in the standard form.
is a standard form
(iv) is not in the standard form.
is standard form.
CHECK YOUR PROGRESS 6.3
1. Solve the following equations using factor method.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Solution: (i) We have, 
or 
Therefore, 
and 
are solutions of the equation.
(ii) We have, 
or 
Therefore, 
and 
are solutions of the equation.
(iii) We have,
or 
Therefore, 
and 
are solutions of the equation.
(iv) We have,
or 
Therefore, 
and 
are solutions of the equation.
(v) We have,
or 
Therefore, 
and 
are solutions of the equation.
(vi) We have,
or 
Therefore, 
and 
are solutions of the equation.
CHECK YOUR PROGRESS 6.4
1. Without determining the roots, comment on nature of roots of following equations:
(i) 
(ii) 
(iii) 
(iv) 
Solution: (i) We have,
Here, 
Therefore, the equation has two real distinct roots.
(ii) We have,
Here, 
Therefore, the equation has two equal roots.
(iii) We have, 
Here, 
Therefore, the equation has two equal roots.
(iv) We have,
Here, 
Therefore, the equation does not have any real root.
2. Solve the following equations using quadratic formula:
(i) 
(ii) 
(iii) 
Solution: (i) We have, 
Here, 
Using quadratic formula, 
Thus, the two roots are 
.
(ii) We have,
Here, 
Using quadratic formula,
Thus, the two roots are 5 and 0 .
(iii) We have, 
Here, 
Using quadratic formula,
Thus, the two roots are 10 and 5 .
3. Find the value of m so that the following equations have equal roots:
(i) 
(ii) 
(iii) 
(iv) 
Solution: (i) We have,
Here, 
A/Q, 
Therefore, the value of m is 
.
(ii) We have,
Here, 
A/Q,
(iii) We have,
Here, 
A/Q,
(iv) We have,
Here, 
A/Q,
CHECK YOUR PROGRESS 6.5
1. The sum of the squares of two consecutive even natural numbers is 164. Find the numbers.
Solution: Let, 
be the two consecutive even natural numbers .
A/Q, 
(impossible) or 
Therefore, the two consecutive even natural numbers are 8 and 10 (=8+2) .
2. The length of a rectangular garden is 7 m more than its breadth. If area of the garden is 144 m² , find the length and breadth of the garden.
Solution: Let, x and y be the length and breadth of the rectangular garden respectively .
A/Q, 
(i)
And 
or 
Putting 
in equation (i) , we get 
Therefore, the length and breadth of the rectangular garden are 9 m and 16 m respectively .
3. The sum of digits of a two digit number is 13. If sum of their squares is 89, find the number.
Solution: Let the digit at ten's place be x and digit at unit's place be y .
The origin number 
A/Q, 
(i)
And 
or 
Putting 
in equation (i) 
Putting 
in equation (i) 
Therefore, the origin number 
or 
.
4. The digit at ten's place of a two digit number is 2 more than twice the digit at unit's place. If product of digits is 24, find the two digit number.
Solution: Let the digit at ten's place be x and digit at unit's place be y .
Therefore, number .
A/Q, 
(i)
And 
(Impossible) or 
Putting 
in equation (i) , we get 
Therefore, the number 
.
5. The sum of two numbers is 15. If sum of their reciprocals is 
, find the two numbers.
Solution: Let one number be x and the other number 
.
A/Q, 
or 
If one number is 10 , then the other number is 5 (= 15 – 10) .
If one number is 5 , then the other number is 10 (= 15 – 5) .
Therefore, the required numbers are 5 and 10 .
TERMINAL EXERCISE
1. Which of the following are quadratic equations?
(i) 
(ii) 
(iii) 
(iv) 
Solution: (i) 
is a quadratic equation .
(ii) is a quadratic equation ,because 
is not quadratic polynomial.
(iii) We have,
is a quadratic equation .
(iv) We have, 
is a quadratic equation .
2. Solve the following equations by factorisation method:
(i) 
(ii) 
(iii) 
(iv) 
Solution: (i) We have, 
or
Therefore, the roots are 8 and – 4 .
(ii) We have, 
or 
Therefore, the roots are 
.
(iii) We have, 
or
Therefore, the roots are – 6 and 3 .
(iv) We have, 
or 
Therefore, the roots are 
.
3. Find the value of m for which 
has equal roots.
Solution: We have, 
Here, 
A/Q,
Therefore, the value of m is 
.
4. Find the value of m for which 
has equal roots.
Solution: We have, 
Here, 
A/Q,
5. Solve the following quadratic equations using quadratic formula:
(i) 
(ii) 
(iii) 
(iv) 
Solution: (i)
Here, 
Using quadratic formula, 
Therefore, the roots are 
(ii) 
Here,
Using quadratic formula,
(iii) We have, 
Here,
Using quadratic formula,
(iv)
Here, 
Using quadratic formula, 
6. The sides of a right angled triangle are x – 1, x and x + 1. Find the value of x and hence the sides of the triangle.
Solution: Since, the sides of a right angled triangle are x – 1, x and x + 1.
Using Pythagoras, we get 
Therefore, the sides of the triangle are 4 – 1 , 4 and 4 +1 ; i.e., 3 , 4 and 5 .
7. The sum of squares of two consecutive odd integers is 290. Find the integers.
Solution: Let x and x+2 be the two consecutive odd integers respectively .
A/Q, 
(Impossible) or 
Therefore, the two consecutive odd integers are 11 and 13 (=11+2) respectively .
8. The hypotenuse of a right angled triangle is 13 cm. If the difference of remaining two sides is 7 cm, find the remaining two sides.
Solution: Let, x and y (x>y) are the two shorter sides of the right angled triangle (in cm) respectively.
A/Q, 
(i)
And 
(Impossible) or 
Putting 
in equation (i) , we get 
Therefore, the remaining two sides are 5 cm and 12 cm .
9. The sum of the areas of two squares is 41 cm². If the sum of their perimeters is 36 cm, find the sides of the two squares.
Solution: Let x and y (x >y ) are the sides of the two squares respectively (in cm) .
A/Q, 
(i)
And 
or 
Putting in equation (i) , we get 
Putting 
in equation (i) , we get 
Therefore, the sides of the two squares are 5 cm and 4 cm .
10. A right angled isosceles triangle is inscribed in a circle of radius 5 cm. Find the sides of the triangle.
Solution: Let each equal leg be x cm.
The diameter = 2 × 5 = 10 cm
By Pythagoras' theorem, 
(only positive value)
Therefore, the sides of the triangle are 
cm , cm and 10 cm .
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