CBSE Class 10 Mathematics Chapter 4 - Quadratic Equations - Exercise 4.3

 CBSE Class 10 Mathematics  Chapter 4 - Quadratic Equations - Exercise 4.3

Q1.	Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i)	2x2 – 3x + 5 = 0
Solution.	To Solve this type of questions, basic concept is: A Standard quadratic equation is ax2 + bx + c = 0 If D < 0, No real root If D = 0, Two equal real roots (coincident roots) If D > 0. Two distinct real roots Where D = b2—4ac Given equation 2x2 – 3x + 5 = 0 Compare equation with ax2 + bx + c = 0 a = 2 , b = —3 and c = 5 We know D = b2—4ac D = (—3)2 — 4(2)(5) D = 9 — 40 D = —31 Because D < 0, so there are no real roots.

(ii)

3x2 – 4√3x + 4 = 0

Solution.

To Solve this type of questions, basic concept is: A Standard quadratic equation is ax2 + bx + c = 0 If D < 0, No real root If D = 0, Two equal real roots (coincident roots) If D > 0. Two distinct real roots Where D = b2—4ac Given equation 3x2 – 4√3x + 4 = 0 Compare equation with ax2 + bx + c = 0 a = 3 , b = —4√3 and c = 4 We know D = b2—4ac D = (—4√3)2 — 4(3)(4) D = (—4√3 × —4√3) — 4(3)(4) D = (—4 × —4 × √3 × √3) — 4(3)(4) D = (16× 3) — 4(3)(4) D = 48 — 48 D = 0 Because D = 0, so there are Two equal real roots.

(iii)

2x2– 6x + 3 = 0

Solution.

To Solve this type of questions, basic concept is: A Standard quadratic equation is ax2 + bx + c = 0 If D < 0, No real root If D = 0, Two equal real roots (coincident roots) If D > 0. Two distinct real roots Where D = b2—4ac Given equation 2x2– 6x + 3 = 0 Compare equation with ax2 + bx + c = 0 a = 2 , b = —6  and c = 3 We know D = b2—4ac D = (—6)2 — 4(2)(3) D = (—6) ×( —6) — 4(2)(3) D = 36 — 24 D = 12 Because D > 0, so there are Two distinct real roots.

Q2.	Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i)	2x2 + kx + 3 = 0
Soution.	2x2 + kx + 3 = 0 Compare equation with ax2 + bx + c = 0 a = 2 , b = k and c = 3 As per question equation has two equal root. So D = 0 b2—4ac = 0

(ii)

kx (x – 2) + 6 = 0

Solution.

kx (x – 2) + 6 = 0 kx2 – 2kx + 6 = 0 ………………. (i) Compare equation with ax2 + bx + c = 0 a = k , b = –2k and c = 6 As per question equation has two equal root. So D = 0 b2—4ac = 0 putting value  (–2k)2 — 4(k)(6) = 0 4k2 — 24k = 0 4k(k — 6) = 0 Now, set each factor equal to zero and solve for k: 4k = 0 K = 0/4 = 0 k — 6 = 0 k = 6 so, k = 0 and k = 6 If k = 0 The equation (i) becomes (0)x2 – 2(0)x + 6 = 0 0 — 0 + 6 = 0 6 = 0 This is not correct. So k = 0 is not possible. If k = 6 The equation (i) becomes (6)x2 – 2(6)x + 6 = 0 6x2 – 12x + 6 = 0 So, k = 6 is possible. To find value of root - 6x2 – 12x + 6 = 0 6x2 – 6x – 6x + 6 = 0 6x(x – 1) – 6(x – 1) = 0 ∴ x = 1

Q3.

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution.

According to question – Area of rectangular mango grove = 800 m2 And length is twice its breadth. Let breadth  = x Then length = 2x We know that – Area of rectangle = length × breadth Putting values – 800 = 2x × x 800 = 2x2 2x2 = 800 x2 = 800/2 x2 = 400 x = ±√400 x = ±20 So, x must be 20 or —20. Because x is breadth and it cannot be negative. So breadth of rectangular mango grove = 20 m Length of rectangular mango grove = 2x = 40 m

Q4.

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution.

According to question – Sum of the age of two friends  = 20 years Let First friend present age = x Then Present age of second friend  = 20 — x Four year ago their age – Ist friend age  = x — 4 2nd friend age = (20—x) — 4 = 16—x Four year ago, product of their ages = 48 years (x — 4)( 16—x) = 48 x( 16—x) — 4( 16—x) = 48  16x—x2 — 64 + 4x = 48 16x—x2 — 64 + 4x — 48 = 0 —x2 + 16x + 4x — 64— 48 = 0 —x2 + 20x  — 112 = 0 x2 — 20x  + 112 = 0 Compare equation with ax2 + bx + c = 0 a = 1 , b = –20 and c = 112 We know D = b2—4ac D = (—20)2 — 4(1)(112) D = (—20) ×( —20) — 4(1)(112) D = 400 — 448 D = —48 Because D < 0, so there are No real roots. Hence, given situation is not possible.

Q5.

Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

Solution.

According to question – Perimeter of rectangular park = 80 m Area of rectangular park = 400 m2 Let length of park = x We know that – 2(length + Breadth) = Perimeter 2(length + Breadth) = 80 length + Breadth = 40 Breadth = 40 — length = 40—x We know that – Length × breadth = area Putting value (x)( 40—x) = 400 40x—x2 = 400 40x—x2 — 400 = 0 —x2 + 40x — 400 = 0 x2 — 40x + 400 = 0 Compare equation with ax2 + bx + c = 0 a = 1 , b = –40 and c = 400 We know D = b2—4ac D = (—40)2 — 4(1)(400) D = (—40) ×( —40) — 4(1)(400) D = 1600 — 1600 D = 0 Because D = 0, so there are Two equal real roots and it is possible to design a rectangular park by given data.