Before talking about the NCERT solutions for Class 10 Maths chapter 4 exercise 4.1, letâ€™s do a quick recapitulation- from chapter Polynomials, we already know what quadratic polynomials are. Isnâ€™t it? When a polynomial of quadratic order of the form asÂ ax2 + bx + c = 0 (a â‰ 0) becomes zero, we termed it as a quadratic equation. Now ax2 + bx + c = 0 (a â‰ 0) is known as the general form of the quadratic equation where â€˜aâ€™ and â€˜bâ€™ are called variables of the first-order and c is a constant. Maths class 10 chapter 4 exercise 4.1 puts students into the practice of solving quadratic equations. And class 10 maths chapter 4 exercise 4.1 solutions lend them a helping hand and make them understand the right method of solving those problems. Students can now download free Chapter 4 Exercise 4.1 NCERT Book Solutions PDF for Class 10 to boost their exam preparation. Vedantu provides NCERT Solutions for Class 10 Science as well which will help students to prepare for their science exam.
The history of the solution of quadratic equations extends across the world for more than 4000 years as class 10 Math chapter 4 exercise 4.1 will tell you. The earliest known records are Babylonian clay tablets from about 1600 BCE where the diagonal of a unit square is given to five decimal places of accuracy. But, as you will know from class 10 quadratic equation exercise 4.1, the use of algebraic symbols only began in the 15th century. Exercise 4.1 Maths class 10 will also inform you that Euclid, a famous Greek mathematician invented the quadratic equations and its solutions while calculating dimensions such as length; breadth and height via geometry. Later Sridharacharya made the calculations of these quadratic equations by completing the square methodology which became more convenient; better known as now as Sridharacharya Formula of Quadratic Equation.
The root of the quadratic equation ax2 + bx + c = 0 (a â‰ 0)Â is given in NCERT class 10 Maths chapter 4 exercise 4.1 by the following formula:
\[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]
This formula is known as Sridharacharya formula, as you will know from exercise 4.1 class 10 Maths.
Here, the quantity Â Â b2 âˆ’4acÂ Â Â is called theÂ discriminantÂ of the polynomial.
According to Sridharacharya, as mentioned in NCERT Maths class 10 chapter 4 exercise 4.1, every quadratic equation has two roots and the types depend on the value of this discriminant.
If b2âˆ’4acÂ Â > 0, the roots of the quadratic equation are real and distinct in nature.
If b2âˆ’4acÂ Â < 0, the roots of the quadratic equation are unreal i.e. has a complex solution.
If b2âˆ’4acÂ Â Â = 0, the roots of the quadratic equation are real and equal.
Example 1:
Examine the nature of the roots of the following quadratic equation.
3x2Â + 8x + 4Â =Â 0.
Solution:
The given quadratic equation is in the general form:
ax2Â + bx + cÂ =Â 0
Then, we have aÂ =Â 3, bÂ =Â 8 and cÂ =Â 4.
Therefore:Â
b2Â - 4acÂ =Â 82Â - 4(3)(4)
b2Â - 4acÂ =Â 64 - 48
b2Â - 4acÂ =Â 16
Here, b2Â - 4ac > 0 and also a perfect square.
So, the roots are real and distinct.
Example 2:
Examine the nature of the roots of the following quadratic equation.
5x2Â - 4x + 2Â =Â 0
Solution:
The given quadratic equation is in the general form:
ax2Â + bx + cÂ =Â 0
Then, we have aÂ =Â 5, bÂ =Â -4 and cÂ =Â 2.
Therefore:
b2Â - 4acÂ =Â (-4)2Â - 4(5)(2)
b2Â - 4acÂ =Â 16 - 40
b2Â - 4acÂ =Â -24
Here, b2Â - 4ac < 0.
So, the roots are imaginary.
A quadratic equation may be solved either by factorizing the left side( when the right side is zero) or by completing a square on the left side.
Example 1:Â SolveÂ 6x2 + 11x - 10 = 0
Solution:Â 6x2 + 11x - 10 = 0
â‡’ 6x2 + 15x - 4x - 10 = 0
â‡’ 3x(2x + 5) - 2(2x + 5) = 0
â‡’ (2x + 5)(3x - 5) = 0
Therefore, eitherÂ 2x + 5 = 0Â or,Â 3x - 5 = 0
Hence,Â x = -5/2 or x = â…”.
Sometimes the equations given are not in quadratic order and hence we need to convert it into a quadratic form to solve it.
Example 1:Â SolveÂ \[2^{2x + 3} + 2^{x + 3} = 1 + 2^{x}\]
Solution:Â \[2^{2x + 3} + 2^{x + 3} = 1 + 2^{x}\]
or,Â \[2^{2x} . 2^{3} + 2^{x} . 2^{3} = 1 + 2^{x}\]
or,Â \[8 . 2^{2x} + 8 . 2^{x} = 1 + 2^{x}\]â€¦(1)
Now, letÂ \[2^{x} = y\]Â Then,Â \[y^{2} = (2^{x})^{2} = 2^{2x}\]
âˆ´Â from (1) we get,
\[8y^{2} + 8y = 1 + y\]
or,Â \[8y^{2} + 7y - 1 = 0\]
or,Â \[8y^{2} - 8y - y - 1 = 0\]
or,Â \[8y(y + 1) - 1(y + 1) = 0\]
or,Â \[(y + 1)(8y - 1) = 0\]
âˆ´Â y = -1
or â…›
If,Â y = -1, then,Â \[2^{x} = -1\], which is not possible.
If,Â y = â…›, thenÂ \[2^{x} = \frac{1}{8}\]
or,Â \[2^{x} = \frac{1}{2^{3}} = 2^{-3}\]
âˆ´Â x = -3.
Example 2:Â SolveÂ \[\sqrt{\frac{x}{1 - x}} + \sqrt{\frac{1 - x}{x}} = \frac{13}{6}\]
Solution:Â Let us putÂ \[y = \sqrt{\frac{1}{1-x}}\], thenÂ \[\frac{1}{y} = \sqrt{\frac{1 - x}{x}}\]
\[\therefore y + \frac{1}{y} = \frac{13}{6} or, \frac{y^{2} + 1}{y} = \frac{13}{6}\]
â‡’ 6y2 + 6 = 13y
â‡’ 6y2 - 13y + 6 = 0
â‡’ 6y2 - 4y - 9y + 6 = 0
â‡’ (3y - 2)(2y - 3) = 0
âˆ´ y = â…” or, y = 3/2
IfÂ y = â…” , then \[\sqrt{\frac{x}{1-x}} = \frac{2}{3}\]
\[\Rightarrow \therefore \frac{x}{1 - x} = \frac{4}{9}\]
â‡’ 9x = 4 - 4x
â‡’ 13x = 4
âˆ´ x = 4/13
IfÂ y = 3/2, then \[\sqrt{\frac{x}{1-x}} = \frac{3}{2}\]
\[\Rightarrow \therefore \frac{x}{1 - x} = \frac{9}{4}\]
â‡’ 4x = 9 - 9x
â‡’ 13x = 9
âˆ´ x = 9/13
If Î± and Î²Â be the roots of the quadratic equationÂ ax2 + bx + c = 0 (a â‰ 0)Â then,
\[\alpha + \beta = -\frac{b}{a}\] is called the sum of roots and \[\alpha \beta = \frac{c}{a}\] is called products of roots of the quadratic equations.
Example 1: If the roots of the equationÂ x2 + px + 7 = 0Â are denoted byÂ Î± and Î²Â andÂ Î±2 + Î²2 = 22, find the value of p.
Solution:Â \[\alpha + \beta = \frac{-p}{1} = (-p)\]Â andÂ \[\alpha \beta = \frac{7}{1} = 7\]
Now,Â Î±2 + Î²2 = 22Â (given)
or,Â (Î± + Î²)2 - 2Î±Î² = 22
or,Â (-p)2 - 2(7) = 22
or,Â p2 = 22 + 14 = 36
p = Â±6
Class 10th Maths Chapter 4 exercise 4.1 has several problems for you to solve. Following are a few of them.
Example - 1: In a class test, the sum of Rekhaâ€™s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution: with the help of class 10 Maths exercise 4.1, let us marks in Maths asÂ xx,
Then as per the problem,
Marks in English would be =Â 30 âˆ’ x30 âˆ’ x.
Now if Rekha got 2 marks less, then Maths marks would =Â (x âˆ’ 2)(x âˆ’ 2)
and if she got 3 marks less in English, then English marks would =Â (30 âˆ’ x âˆ’ 3) = (27 âˆ’ x)(30 âˆ’ x âˆ’ 3) = (27 âˆ’ x)
As per problem, the product is equal to 210 so
(x âˆ’ 2)(27 âˆ’ x) = 120(x âˆ’ 2)(27 âˆ’ x) = 120
âˆ’x2 + 25x + 54 = 210 âˆ’ x2 + 25x + 54 = 210
or
x2 âˆ’ 25 x + 156 = 0 x 2 âˆ’ 25 x +156 = 0
x2 âˆ’ 90x + 30x âˆ’ 2700 = 0 x 2 âˆ’ 90x + 30x âˆ’ 2700 = 0
or x = 12 or 13
So Rekhaâ€™ marks in Maths is 12 or 13 and the English marks are 18 or 17.
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1.Â The Altitude of a Right Triangle is 7 cm Less than its Base. If the Hypotenuse is 13 cm, Find the Other Two Sides.
Answer: You can find the following in class 10 Maths ex 4.1 solutions:
Let x be the base, the altitude will be x - 7
Now as per Pythagoras theorem,
x^{2} + (x âˆ’ 7)2 = 13^{2}
= x^{2} + x^{2} - 14x + 49 = 169
= 2x^{2} -14x = 120
= x^{2} - 7x = 60
= x^{2} - 12x + 5x - 60 = 0
= x(x - 12) + 5(x - 12) = 0
= (x - 12)(x + 5) = 0
x = 12 or -5
Since x cannot be negative x = 12.
So base is 12 cm and altitude is (12 - 7) = 5 cm.
2.Â If Zeba were Younger by 5 years than what she really is then the Square of her age Would have been 11 more than five times her actual age. What is her age now?
Answer: The solution according to exercise 4.1 class 10 Maths NCERT solutions is as follows:
Let age of Zeba be x years
As per the question:
(x âˆ’ 5)^{2} = 11 + 5x(x âˆ’ 5)^{2} = 11 + 5x
=> x^{2} + 25 âˆ’ 10x = 11 + 5x(x)^{2} + 25 âˆ’ 10x = 11 + 5x
=> x^{2} âˆ’ 15x + 14 = 0 x 2 âˆ’ 15x + 14 = 0Â x 2 âˆ’ 14x âˆ’ x + 14 = 0 x 2 âˆ’ 14x âˆ’ x + 14 = 0
=> (x - 1)(x - 14)
Therefore, age of Zeba is 14 years.
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