Math Assignment Class X Ch-4 | Quadratic Equations
Math Assignment Class X
Chapter 4 Quadratic Equations
Extra questions of chapter 4 class 10 Quadratic Equations with answer and hints to the difficult questions. Important and useful math assignment for the students of class 10
ASSIGNMENT FOR 10 STANDARD QUADRATIC EQUATIONS
Question 1. Check whether the given values of x are the solution of equation or not.
a) x2 + √2x – 4 = 0 : At x = √2, x = -2√2 : Ans [yes]
b) (2x + 3)(3x - 2) = 0 : At x = 2/3 : Ans [yes]
c) 6x2 – x – 1 = 0 : At x = 1/2, x = 3/2
Ans
[At x=1/2 yes, at x= 3/2 no]
Question 2. Find k if the value of x is the solution of Q. E.
a) 3x2 - 2kx + 5 = 0 : At x = 2
Ans
[k = 17/4]
b) x2 - (a + b)x + k = 0 : At x = a
Ans [ k = ab ]
c) kx2 + √3x + 6 = 0 : At x = √3
Ans [ k = - 3]
Question 3. If x = - 2 and x = -1/5 are the solution of 5x2 + k x + p = 0 then find k and p.
Ans [K = 11, P = 2]
Question 4. Check whether 3 is the root of this equation.
a) y2 - 5 = 0 [Ans ± √5 ]
b) (2x + 3)2 = 81 [Ans -6 & 3]
c) ax2 - 2bx = 0 [Ans 0, 2b/a]
d) 4√5x2 + 7x - 3√5 = 0 [Ans -3/√5, √5/4]
e) 5x2 - 17/2 x + 3/2 = 0 [Ans [3/2, 1/5]
f) 3a2x2 + 8abx + 4b2 = 0 [Ans -2b/a, -2b/3a]
g) a2b2x2 + b2x - a2x – 1 = 0 [Ans 1/b2, -1/a2]
h) ax2 + (4a2 - 3b)x - 12ab = 0 [Ans -4a, 3b/a]
i). [Ans 5, 5/2]
Question 6. If - 4 is a root of the equation x2 + px – 4 = 0 and the equation x2 + p x + k = 0 has equal roots. Find k.
Ans [k = 9/4 ]
Question 7. If one root of the equation
Ans [k = 3, other root -1].
Question 8. Find the value of m for which the roots of the equation mx(6x + 10) + 25 = 0 are equal.
Question 9. Discuss the nature of the roots of the following
a) 9x2
-
12x + 4 = 0 b) x2 – 3 x
= 0
c) x2 - 1/3x +3/4 = 0 d) x2 - 2√2x - 2√3 = 0
e) √3x2 - 2√2x - 2√3 = 0 f) √2x2 - √5x + 3 = 0
Question 10. Find the quadratic equation whose roots are
a) -8 & -3 [Ans
x2
+
11x + 24 = 0 ]
b) 1 + √2 and 1 - √2 [Ans x2 - 2x - 1 = 0 ]
c)
Question 11. For what value of p will the following equation have real roots
a) 9x2
-
5x + (p + 1) = 0 [Ans p
b) 2x2 – p x – 4 = 0 [Ans No Value of P]
Question 12. For what value of k the equations have real and repeated roots(or equal roots)
a) (3k + 1)x2
+
2(k + 1)x + k = 0 [ Ans K = -1/2, 1]
b) x2 - 2(k + 1)x + k2 = 0 [Ans k = -1/2]
c) 4x2 - 2(k + 1) + k + 4 = 0 [Ans k = 5, k = - 3]
d) 9x2 + 8kx + 16 =0 [Ans k= ±3]
e) (k-12)x2 + 2(k - 12)x + 2 = 0 [Ans k = 12]
f) (k + 4)x2 + (k +1)x + 1 = 0 [Ans k = 5, -3]
Question 13. Find the roots of the following by using quadratic formula.
a) 3x2 + 2√5x – 5 = 0 Ans [√5/3, - √5]
b) p2x2 + (p2 - q2)x - q2 = 0 : Ans [q2/p2, -1]
c) 12abx2 - (9a2 - 8b2)x - 6ab = 0 : Ans [3a/4b, -2b/3a]
d) (a + b) 2x2 + 8(a2 - b2)x + 16(a - b) 2 = 0 [Ans
e) 4x2 - 4a2x + (a4 - b4) = 0 Ans
f) 9x2 - 9(a + b)x + 2a2 + 5ab + 2b2 = 0 Ans
g) 4x2 - 2(a2 + b2)x + a2b2 = 0 Ans [a2/2 , b2/2]
h)
i) 3y2 + (6 + 4a)y + 8a = 0 Ans [-2, - 4a/3]
j)
Question 14. If 𝞪 and 𝞫 are the roots of the
quadratic equation ax2 + bx + c = 0, then find the value of:
a) 𝞪2 + 𝞫2 [Ans
Hint: Using the identity : a2 +b2 = (a + b)2
– 2ab
b) 𝞪3 + 𝞫3 Ans
Hint : Using the identity : a3 + b3
= (a + b)3 - 3ab(a + b)
c)
Question 15. If
a) 𝞪2 + 𝞫2 - 4 𝞪𝞫 Ans [73/4]
b) 𝞪3 + 𝞫3 Ans[-245/8]
Question 16. Solve: a) √x + 2x =1 Ans (1, 1/4)
b)
Hint: Bring all terms without square root to the RHS. Squaring on both side to get a QE and then solve it.
Ans : x = - a, - b
Hint: Bring 1/x to the LHS
Question 18. Solve: (a)
Ans (-4, - 4/7)
Solution Hint:
Putting and
and find the QE in terms of y
Solve the QE and find the value of y. Find the value of y the replace y into x. Again solve the equation for the value of x.
Higher Order Thinking SkillHOTSQuestion 30. Solve: a)
Ans (1, 2, 1/2)Solution Hint: Putting
then squaring on both sides and find the value of
Now find and solve the QE in terms of y
Now replacing the value of y in terms of x
Solve b)
Ans (-1/2, 2, -1/4, 4)
Solution Hint: Putting
then squaring on both sides and find the value of
Now find and solve the QE in terms of y
Now replacing the value of y in terms of x
Question 31. Out of a number of Saras birds, (1/4)th of the number are moving about in Lotus plants, (1/9)th coupled with (1/4)th as well as 7 times the square root of the number move on a hill, 56 birds remain on the trees. What is the total no. of words. Ans [576]
Solution 31Let total birds = x
No. of birds moving in lotus plant = (1/4)x
No. of birds coupled with = (1/9 +1/4)x = (13/36)x
No of birds moving on the hill = 7x
No. of birds on the tree = 56
Acoording to the question







Putting 


+6(y-24)=0)
(y+6)=0)


^{2}=576)
Hence Total Number of birds = 576
Question 32.7 years ago, age of Varun was five times the square of the age of Swati. After 3 years, age of Swati will be 2/5 of the age of Varun. Find their present ages.Solution :
Let 7 years ago the age of Swati = x years
Let 7 years ago the age of Varun = y years
ATQ y = 5x2 …………………(1)
Present age of Swati = x + 7
Present age of Varun = y + 5
After 3 years age of Swati = x + 10
After 3 years age of Varun= y + 10
Again ATQ : Age of Swati = (2/5) age of Varun
x + 10 = (2/5)[y + 10]
5(x + 10) = 2y + 20
5x + 50 = 2y + 20
Putting y = 5x2 we get
5x + 50 = 2(5x2 ) + 20
10x2 - 5x + 20 - 50 = 0
10x2 - 5x - 30 = 0
2x2 - x - 6 = 0 (Dividing by 5)
(2x + 3)(x - 2) = 0
x = - 3/2 (Rejected) and x = 2
Putting x = 2 in equation (1) we get
y = 5 (2)2 ⇒ y = 20
Present age of Swati = x + 7 = 2 + 7 = 9
Present age of Varun = y + 7 = 20 + 7 = 27
Ans (1, 2, 1/2)
Putting then squaring on both sides and find the value of
Now find and solve the QE in terms of y
Now replacing the value of y in terms of x
Solve b)
Ans (-1/2, 2, -1/4, 4)
Solution Hint:
Putting then squaring on both sides and find the value of
Now find and solve the QE in terms of y
Now replacing the value of y in terms of x
Solution 31
Let total birds = x
No. of birds moving in lotus plant = (1/4)x
No. of birds coupled with = (1/9 +1/4)x = (13/36)x
No of birds moving on the hill = 7x
No. of birds on the tree = 56
Acoording to the question
Putting
Hence Total Number of birds = 576
Question 32.
Solution :
Let 7 years ago the age of Swati = x years
Let 7 years ago the age of Varun = y years
ATQ y = 5x2 …………………(1)
Present age of Swati = x + 7
Present age of Varun = y + 5
After 3 years age of Swati = x + 10
After 3 years age of Varun= y + 10
Again ATQ : Age of Swati = (2/5) age of Varun
x + 10 = (2/5)[y + 10]
5(x + 10) = 2y + 20
5x + 50 = 2y + 20
Putting y = 5x2 we get
5x + 50 = 2(5x2 ) + 20
10x2 - 5x + 20 - 50 = 0
10x2 - 5x - 30 = 0
2x2 - x - 6 = 0 (Dividing by 5)
(2x + 3)(x - 2) = 0
x = - 3/2 (Rejected) and x = 2
Putting x = 2 in equation (1) we get
y = 5 (2)2 ⇒ y = 20
Present age of Swati = x + 7 = 2 + 7 = 9
Present age of Varun = y + 7 = 20 + 7 = 27


good collection of questions
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