quadratic-equationsWHERE cd.courseId=9 AND cd.subId=6 AND chapterSlug='quadratic-equations' and status=1SELECT ex_no,page_number,question,question_no,id,chapter,solution FROM question_mgmt as q WHERE courseId='9' AND subId='6' AND chapterId='271' AND ex_no!=0 AND status=1 ORDER BY ex_no,CAST(question_no AS UNSIGNED) CBSE Class 10 Free NCERT Book Solution for Mathematics

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Chapter 4 : Quadratic Equations


At Saralstudy, we are providing you with the solution of Class 10 Mathematics Quadratic Equations according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 10 Mathematics Quadratic Equations so that you can prepare for the exam according to your own pace and your speed.

Exercise 1 ( Page No. : 73 )
Q:
A:

Exercise 1 ( Page No. : 73 )


Exercise 2 ( Page No. : 76 )

Exercise 2 ( Page No. : 76 )
Q:
A:

(i) let x and y are the no of marbles of Romil and Somi respectively.

Acc. to ques,

X + y = 45................... (1)

When they both lost their 5 marbles

(x – 5) (y – 5) = 124.................... (2)

From equation (1), we have

y = 45 – x

Put this value of x in equation (2), we get

(x – 5)(45 – x – 5) = 124

(x – 5)(40 – x) = 124

40x – x2 – 200 + 5x = 128

​x2 - 45x + 324 = 0

x2 - 36x - 9x + 324 = 0

x (x – 36) – 9 (x – 36) = 0

(x – 9) (x – 36) = 0

x = 9 or x = 36


Exercise 2 ( Page No. : 76 )

Exercise 2 ( Page No. : 76 )

Exercise 2 ( Page No. : 76 )

Exercise 2 ( Page No. : 76 )

Exercise 3 ( Page No. : 88 )
Q:
A:

Exercise 3 ( Page No. : 88 )


Exercise 3 ( Page No. : 88 )

Exercise 3 ( Page No. : 88 )

Exercise 3 ( Page No. : 88 )

Exercise 3 ( Page No. : 88 )
Q:
A:

Let the shorter side of rectangular field be = x meter

The diagonal of the field = (x + 60) m

And, longer side is 30m more than shorter side = (x + 30) m

According to question;

Using Pythagoras theorem;

(x + 60)2 = x2 + (x + 30)2

x (x - 90) + 30 (x - 90) = 0

(x + 30) (x - 90) = 0

(x + 30) = 0 or (x - 90) = 0

Either, x = -30 or x = 90

But x ≠ -30 because length can never be negative.

Hence, shorter side of field is 90 m and longer side is (x + 30) = 120 m


Exercise 3 ( Page No. : 88 )

Exercise 3 ( Page No. : 88 )

Exercise 3 ( Page No. : 88 )

Exercise 3 ( Page No. : 88 )

Exercise 3 ( Page No. : 88 )

Exercise 4 ( Page No. : 91 )
Q:
A:

(i)                 2x2 – 3x + 5 = 0

On comparing given equation ax2 + bx + c = 0

We get,

a = 2, b = -3 and c = 5

We know discriminant (D) = b2 – 4ac

D = (-3)2 – 4   2   5

D = 9 – 40

D = -31

As, b2 – 4ac < 0

Therefore, no real roots exist for the given equation.


Exercise 4 ( Page No. : 91 )

Exercise 4 ( Page No. : 91 )

Exercise 4 ( Page No. : 91 )

Exercise 4 ( Page No. : 91 )