CBSE Class 10 Mathematics Quadratic Equation Worksheet Set D

Read and download free pdf of CBSE Class 10 Mathematics Quadratic Equation Worksheet Set D. Students and teachers of Class 10 Mathematics can get free printable Worksheets for Class 10 Mathematics Chapter 4 Quadratic Equation in PDF format prepared as per the latest syllabus and examination pattern in your schools. Class 10 students should practice questions and answers given here for Mathematics in Class 10 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 10 Mathematics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems with solutions on daily basis to get more score in school exams and tests

Worksheet for Class 10 Mathematics Chapter 4 Quadratic Equation

Class 10 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 4 Quadratic Equation in Class 10. This test paper with questions and answers for Class 10 will be very useful for exams and help you to score good marks

Class 10 Mathematics Worksheet for Chapter 4 Quadratic Equation

 

QUADRATIC EQUATIONS

Q.- Without solving, examine the nature of roots of the equations :
(i) 2x2 + 2x + 3 = 0
(ii) 2x2 – 7x + 3 = 0
(iii) x2 – 5x – 2 = 0
(iv) 4x2 – 4x + 1 = 0
 
Sol. (i) Comparing 2x2 + 2x + 3 = 0
with ax2 + bx + c = 0; we get : a = 2, b = 2 and c = 3
D = b2 – 4ac = (2)2 – 4 × 2 × 3 = 4 – 24
= – 20; which is negative.
∴The roots of the given equation are imaginary.
 
(ii) Comparing 2x2 – 7x + 3 = 0
with ax2 + bx + c = 0;
we get : a = 2, b = – 7 and c = 3
D = b2 – 4ac = (–7)2 – 4 × 2 × 3
= 49 – 24 = 25, which is perfect square.
∴The roots of the given equation are rational and unequal.
 
(iii) Comparing x2 – 5x – 2 = 0
with ax2 + bx + c = 0;
we get : a = 1, b = – 5 and c = – 2
D = b2 – 4ac = (–5)2 – 4 × 1 × – 2
= 25 + 8 = 33 ; which is positive but not a perfect square.
∴The roots of the given equation are irrational and unequal.
 
(iv) Comparing 4x2 – 4x + 1 = 0
with ax2 + bx + c = 0;
we get : a = 4, b = – 4, and c = 1
D = b2 – 4ac = (–4)2 – 4 × 4 × 1
= 16 – 16 = 0
∴ Roots are real and equal
 
Q.- For what value of m, are the roots of the equation (3m + 1) x2 + (11 + m) x + 9 = 0 equal?
 
Sol. Comparing the given equation
with ax2 + bx + c = 0;
we get : a = 3m + 1, b = 11 + m and c = 9
∴  Discriminant, D = b2 – 4ac
= (11 + m)2 – 4(3m + 1) × 9
= 121 + 22m + m2 – 108 m – 36
= m2 – 86m + 85
= m2 – 85m – m + 85
= m(m – 85) – 1 (m – 85)
= (m – 85) (m – 1)
Since the roots are equal, D = 0
=> (m – 85) (m – 1) = 0
=> m – 85 = 0 or m – 1 = 0
=> m = 85 or m = 1 
 
Q.- If one of the roots of the quadratic equation 2x2 + px + 4 = 0 is 2, find the the value of p. also find the value of the other roots.
 
Sol. As, 2 is one of the roots, x = 2 will satisfy the
equation 2x2 + px + 4 = 0
=> 2(2)2 + p(2) + 4 = 0
=> 8 + 2p + 4 = 0
i.e., 2p = – 12 and p = – 6
Substituting p = – 6 in the equation
2x2 + px + 4 = 0; we get : 2x2 – 6x + 4 = 0
=> x2 – 3x + 2 = 0
[Dividing each term by 2]
=> x2 – 2x – x + 2 = 0
=> x(x – 2) (x – 1) = 0
=> x – 2 = 0        or x – 1 = 0
=> x = 2             or x = 1
∴ The other (second) root is 1.
 
Q.- In the following, find the value (s) of p so that the given equation has equal roots.
(i) 3x2– 5x + p = 0
(ii) 2px2 – 8x + p = 0

Sol. (i) Comparing 3x2 – 5x + p = 0

with ax2 + bx + c = 0,

we get : a = 3, b = – 5 and c = p
Since, the roots are equal ; the discriminant
b2 – 4ac = 0
i.e., (–5)2 – 4 × 3 × p = 0
=> 25 – 12p = 0 and p = 25/12  = 2 1/12
 
(ii) Comparing 2px2 – 8x + p = 0
with ax2 + bx + c = 0;
we get : a = 2p, b = – 8 and c = p
b2 – 4ac = 0 [Given, that the roots are equal]
=> (–8)2 – 4 × 2p × p = 0
=> 64 – 8p2 = 0
=> – 8p2 = – 64, p2 = 8 and p = ± √8
i.e., p = ± 2√ 2
 
Q.- If α and β are the roots of the quadratic equation ax2 + bx + c = 0, (a ≠ 0) then find the values of :
(i) α2 + β2
(ii) α3 + β3
(iii)α /β + β/α
 

Quadratic equations notes 1

Quadratic equations notes 2

Q.- Solve the following equations :
(i) x4 – 26x2 + 25 = 0
(ii) z4 – 10z2 + 9 = 0
 
Sol. (i) Substituting x2 = y :
x4 – 26x2 + 25 = 0
=> y2 – 26y + 25 = 0
i.e., y2 – 25y – y + 25 = 0
=> y(y – 25) – 1(y – 25) = 0
i.e., (y – 25) (y – 1) = 0
=> y – 25 = 0 or y – 1 = 0
i.e., y = 25 or y = 1
y = 25 => x2 = 25 | y = 1
=> x2 = 1
=> x = ± 5 |
=> x = ± 1
∴Roots of the given equation are : ± 5, ± 1

 

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