NCERT full solutions here you can find full detailed solutions for all the questions with proper format. class 10 maths ncert solutions chapter 4 exercise 4.4.
You can see the solution for complete chapter here –
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Complete 10th Class Maths Solution
- NCERT Class 10th – Chapter 3- Exercise 3.6 Pair of Two Equations in Two Variables
- NCERT Class 10th – Chapter 4 – Exercise 4.1 Quadratic Equations
- NCERT Class 10th – Chapter 4 – Exercise 4.2 Quadratic Equations
EXERCISE:-4.4
QUE:-1. Find the nature of the roots of the following quadratic equations. If the real roots exist,
find them:
(i) 2x²– 3x + 5 = 0 ii) 3x² – 4√3 x + 4 = 0
(iii) 2x² – 6x + 3 = 0
SOL:-
A quadratic equation ax²+ bx + c = 0 has
⇒ two distinct real roots, if b² – 4ac > 0,
⇒ two equal roots (i.e., coincident roots), if b² – 4ac = 0, and
⇒ no real roots, if b² – 4ac < 0.
(i) 2x²– 3x + 5 = 0
compairing with ax²+ bx + c = 0
a =2 , b= -3 , c = 5
b² – 4ac = -3² – 4×2 ×5
= 9 -80 =-31 < 0.
therefore the quadratic equation has no real roots.
(ii) 3x² – 4√3 x + 4 = 0
compairing with ax²+ bx + c = 0
a =3 , b= -4√3 , c = 4
b² – 4ac = (-4√3)² – 4×3×4
= 48 -48 =0
if b² – 4ac = 0 then quadratic equation have two equal roots.
the roots are -b/2a and -b/2a
-(-4√3) /2 ×3 =2/√3
therefore roots are 2/√3 and 2/√3 .
(iii) 2x² – 6x + 3 = 0
compairing with ax²+ bx + c = 0
a =2 , b= -6 , c = 3
b² – 4ac = (-6)² – 4×2×3
= 36 -24 =12 > 0,
if b² – 4ac > 0,then quadratic equation has two distinct real roots
x= (-b ±√b² -4ac)/2a
= (-(-6) ±√(-6)² – 4×2×3)/2 ×2
= (6 ± √ 36 -24 )/4
= 2(3 ± √3)/4 =(3 ± √3)/2
so roots are (3 + √3)/2 and (3 – √3)/2
QUE:- 2. Find the values of k for each of the following quadratic equations, so that they have two
equal roots.
(i) 2x² + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
SOL:-
(i) 2x² + kx + 3 = 0
⇒ two equal roots (i.e., coincident roots), b² – 4ac = 0,
compairing with ax²+ bx + c = 0
a =2 , b= k , c = 3
k² – 4×2×3 = 0
k² – 24 =0
k² =24
k =±√24 = ±2√6
(ii) kx (x – 2) + 6 = 0
kx² -2kx +6 = 0
⇒ two equal roots (i.e., coincident roots), b² – 4ac = 0,
compairing with ax²+ bx + c = 0
a =k , b= 2k , c = 6
4k² – 4k×6 = 0
4k² – 24k = 0
4k(k -6) =0
roots are k=0 or k=6
so k = 6
QUE:-3. Is it possible to design a rectangular mango grove whose length is twice its breadth,
and the area is 800 m²? If so, find its length and breadth.
SOL:-
let the length of the rectangular mango grove is x
according to the question breadth of it 2x
given that x × 2x =800
2x² =800
x² =800/2=400
x² -400=0
⇒ ⇒ two distinct real roots, if b² – 4ac > 0,
compairing with ax²+ bx + c = 0
a =1 , b= 0 , c = -400
b² – 4ac = 0 – 4 ×1 × (-400)
= 1600
b² – 4ac > 0
∴ x =±20
length never negetive for a rectangular surface x =20
breadth is 2x = 2×20=40
QUE:-4. Is the following situation possible? If so, determine their present ages.The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
SOL:-
let the age of first friend is x
according to question second friend ‘s age 20-x
age 4 year ago 1st friend x-4
and 2nd friend’s age 20-x-4 =16-x
as given in question
(x-4)(16-x)=48
16x -x² -64 +4x =48
-x² +20x -112 =0
compairing with ax²+ bx + c = 0
a =-1 , b= 20 , c = -112
b² – 4ac = 20² – 4(-1)(-112)
= 400 -448=-48
no real roots for the given situation,because b² – 4ac < 0.
QUE:-5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find
its length and breadth.
SOL:-
let the lenghth of the park is l
and breadth of the park is b
according to question area of park is l×b =400
b =400/l
perimeter of park 2(l +b) =80
l+b =80/2
l + 400/l =40
l² +400=40l
l² +400 -40l=0
compairing with ax²+ bx + c = 0
a = 1 , b = -40 c =400
b² – 4ac = -40² – 4×1×400
= 1600-1600
=0
⇒ two equal roots (i.e., coincident roots), if b² – 4ac = 0,
l= (-b ±√b² -4ac)/2a
= (-(-40) ±√(-40)² -4×1×400)/2×1
= (40 ±√1600-1600)/2
= 40/2=20
l =20
b =400/20 =20
length of park is 20 meter
breadth of park is 20 meter
In this chapter, we have studied about the following points:
1. A quadratic equation in the variable x is of the form ax² + bx + c = 0, where a, b, c are real
numbers and a ¹ 0.
2. A real number a is said to be a root of the quadratic equation ax² + bx + c = 0, if
aα² + bα + c = 0. The zeroes of the quadratic polynomial ax² + bx + c and the roots of the
quadratic equation ax² + bx + c = 0 are the same.
3. If we can factorise ax² + bx + c, a≠ 0, into a product of two linear factors, then the roots
of the quadratic equation ax² + bx + c = 0 can be found by equating each factor to zero.
4. A quadratic equation can also be solved by the method of completing the square.
5. Quadratic formula: The roots of a quadratic equation ax² + bx + c = 0 are given by
(-b ±√b² -4ac)/2a provided b² – 4ac ≥ 0.
6. A quadratic equation ax² + bx + c = 0 has
(i) two distinct real roots, if b² – 4ac > 0,
(ii) two equal roots (i.e., coincident roots), if b² – 4ac = 0, and
(iii) no real roots, if b² – 4ac < 0.
You can see the solution for complete chapter here –
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