CBSE Class 10 Mathematics Chapter 4 - Quadratic Equations - Examples

CBSE Class 10 Mathematics  Chapter 4 - Quadratic Equations - Examples

Example 1.	Represent the following situations mathematically:
(i)	John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Solution.	Given- Total number of marbles = 45 Let the number of marble John have = x Then number of marble Jivanti have = 45 — x After they lost 5 marbles each Number of marble with John  = x — 5 Number of marble with Jivanti = 45 — x — 5 = 40 — x According to question product of marble now = 124 (x — 5)( 40 — x) = 124 x(40 — x) — 5(40 — x)=124 40x — x2 — 200 + 5x = 124 45x — x2 — 200 = 124 0 = 200 + 124 + x2—45x 324 = x2—45x 0 = x2 —45x + 324 x2 —45x + 324 = 0 So required quadratic equation is x2 —45x + 324 = 0 For find out how many marbles they had to start with, Split middle term – ⇒ x2 —36x—9x + 324 = 0 ⇒ x(x —36)—9(x — 36) = 0 ⇒ (x —9)(x — 36) If john marble is 36 then Jivanti marble is (45—36) = 9 If john marble is 9 then Jivanti marble is (45—9) = 36

(ii)

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

Solution.

As we know number of toy produced × production cost of each toy = Total cost of production Let the number of toys produced each day = x Cost of production each toys = Rs. (55—x) It is given that total production of the toys = Rs. 750 x(55—x) = 750 55x — x2 = 750 x2 — 55x + 750 = 0 So required quadratic equation is x2 — 55x + 750 = 0 For find out possible value of toys produced that day, Split middle term – ⇒ x2 —25x—30x + 750 = 0 ⇒ x(x —25)—30(x — 25) = 0 ⇒ (x —25)(x — 30) So, either number of toys (x ) = 25 or 30.

Example 2.	Check whether the following are quadratic equations:
(i)	(x – 2)2 + 1 = 2x – 3
Solution.	We know that the standard form of a quadratic equation is ax2 + bx + c = 0         where a ≠ 0 and the highest power of the variable in the equation should be 2. To check equation writing it in standard form: (x – 2)2 + 1 = 2x – 3 (x2 + 4 – 4x) + 1 = 2x – 3        [ by using (a – b)2 = (a2 + b2 – 2ab)] x2 + 5 – 4x = 2x – 3 x2 + 5 – 4x – 2x + 3 = 0 x2 – 6x  + 8 = 0 Form of this equation like standard form of a quadratic equation where a = 1, b = –6 and c = 8 So it is a quadratic equation.

(ii)

x(x + 1) + 8 = (x + 2) (x – 2)

Solution.

We know that the standard form of a quadratic equation is ax2 + bx + c = 0         where a ≠ 0 and the highest power of the variable in the equation should be 2. To check equation writing it in standard form: x(x + 1) + 8 = (x + 2) (x – 2) x(x + 1) + 8 = x2 – 4        [ by using (a + b) (a – b) = a2 – b2 ] x2 + x + 8 = x2 – 4 x2 + x + 8 – x2 + 4 = 0 (x2 – x2) + x + 8 + 4 = 0 x + 12 = 0 Form of this equation have highest power is 1 not 2, so it is not a quadratic equation.

(iii)

x (2x + 3) = x2 + 1

Solution.

We know that the standard form of a quadratic equation is ax2 + bx + c = 0         where a ≠ 0 and the highest power of the variable in the equation should be 2. To check equation writing it in standard form: x (2x + 3) = x2 + 1 2x2 + 3x = x2 + 1 2x2 + 3x – x2 – 1 = 0 x2 + 3x – 1 = 0 Form of this equation like standard form of a quadratic equation where a = 1, b = 3 and c = –1 So it is a quadratic equation.

(iv)

(x + 2)3 = x3 – 4

Solution.

We know that the standard form of a quadratic equation is ax2 + bx + c = 0         where a ≠ 0 and the highest power of the variable in the equation should be 2. To check equation writing it in standard form: (x + 2)3 = x3 – 4 [ by using (a + b)3 = a3 + b3 + 3a2b + 3ab2] x3 + 23 + 3(x2)(2) + 3(x)(2)2 = x3 – 4 x3 + 8 + 6x2 + 12x = x3 – 4 x3 + 8 + 6x2 + 12x – x3 + 4 = 0 6x2 + 12x + 12 = 0 6(x2 + 2x + 2) = 0 x2 + 2x + 2 = 0 Form of this equation like standard form of a quadratic equation where a = 1, b = 2 and c = 2 So it is a quadratic equation.

Q3.	Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation.
Solution.	Note : for splitting middle term, we need to find two number whose sum and multiply (product) is equal to middle term. 2x2 – 5x + 3 = 0 Split middle term 2x2 – 3x – 2x + 3 = 0 x (2x– 3) – 1 (2x – 3) = 0 (2x– 3)(x – 1) = 0 Now, Set each factor equal to zero and solve x: x – 1 = 0 x = 1 2x – 3 = 0 2x = 3 x = 3/2 So, root of the equation is x = 1 and x = 3/2

Q4.	Find the roots of the quadratic equation 6x2 – x – 2 = 0.
Solution.	Note : for splitting middle term, we need to find two number whose sum and multiply (product) is equal to middle term. 6x2 – x – 2 = 0 Split middle term 6x2 + 3x – 4x – 2 = 0 3x (2x + 1) – 2 (2x + 1) = 0 (3x – 2)(2x + 1) = 0 Now, Set each factor equal to zero and solve x: 3x – 2 = 0 3x = 2 x = ⅔ 2x + 1 = 0 2x = –1 x = – ½ So, root of the equation is x = ⅔ and x = – ½

Q5.	Find the roots of the quadratic equation 3x2 – 2√6x + 2 = 0
Solution.

Q6.

Find the dimensions of the prayer hall discussed in Section 4.1. Suppose a charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall?

Solution.

As shown in figure, charity hall has rectangular size and we know that Area of a rectangle (hall) = length × breadth According to question, Area of hall is 300 m2 And is one meter more than twice its breadth Let breadth of hall = x meters Then length is = (2x + 1) meters So area of hall = length × breadth 300 = x(2x + 1) 300 = 2x2 + x 0 = 2x2 + x – 300 2x2 + x – 300 = 0 Split middle term 2x2 + 25x – 24x – 300 = 0 x (2x + 25) – 12(2x + 25) = 0 (x – 12)(2x + 25) = 0 Now, Set each factor equal to zero and solve x: x – 12 = 0 x = 12 2x + 25 = 0 2x = – 25 x = – 25/2 But x cannot be negative as breadth cannot be negative So, breadth of hall (x) = 12 meters Length of hall = (2x + 1) = 2(12) + 1 = 25 meters

Q7.	Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and hence find the nature of its roots.
Solution.	2x2 – 4x + 3 = 0 By comparing this equation to standard form of a quadratic equation which is ax2 + bx + c = 0       where a ≠ 0 We find a = 2 , b = – 4 and c = 3 As we know discriminant (D) = b2 —4ac Put the value D = (– 4)2 — 4 (2)(3) D = 16 — 24 D = — 8 Because here D < 0 ; equation has no real root.

Q8.

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary are 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Solution.

Let P is the location of pole on the boundary. As shown in figure, AB = Diameter of circular park = 13m According to question difference of the distance of the pole from the gate A and B is 7m. Its mean BP — AP or AP — BP = 7m Let’s take AP — BP = 7m AP = BP + 7m Assume BP = x, then AP = x + 7m We know Angle in a semicircle is a right angle. Since AB is a diameter, so ∠APB = 900 ΔAPB is a right angle triangle, use Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 AB2 = BP2 + AP2 132 = x2 + (x + 7)2 By using (a+b)2 = a2 + b2 + 2ab 169 = x2 + x2 + 72 + 2 (x)(7) 169 = x2 + x2 + 49 + 14x x2 + x2 + 49 + 14x —169 = 0 2x2  + 14x —120 = 0 2 (x2  + 7x —60) = 0 x2  + 7x —60 = 0 by comparing this equation to standard form of a quadratic equation a = 1 , b = 7 and c = —60 As we know discriminant (D) = b2 —4ac Put the value D = (7)2 — 4 (1)(—60) D = 49—4×—60 D = 49 + 240 D = 289

Q9.	Find the discriminant of the equation 3x2 – 2x +⅓= 0 and hence find the nature of its roots. Find them, if they are real.
Solution.	3x2 – 2x +⅓= 0 For removing fraction multiply by 3 3(3x2) – 3(2x) +3(⅓) = 3(0) 9x2 – 6x + 1 = 0 by comparing this equation to standard form of a quadratic equation a = 9, b = —6 and c = 1 D = b2 — 4ac Put value D = (—6)2 — 4 (9)(1) D = 36 — 36 D = 0