Unit # 2 Theory Of Quadratic Equation Exercise 2.1 4ma47

Unit # 2 Theory of Quadratic Equation

Exercise 2.1

Question 1 Find the discriminant of the following given quadratic equation. (iii)

9x2 – 30x + 25 = 0

Solution 9x2 – 30x + 25 = 0 Here, a = 9, b = -30, c = 25 Disc = b2 – 4ac Disc = ( -30)2 – 4 ( 9) ( 25) Disc = 900 – 900 Disc = 0 (iv)

4x2 – 7x -2 = 0

Solution 4x2 – 7x -2 = 0 Here, a = 4 , b = -7 , c = -2 Disc = b2 – 4ac Disc = ( -7)2 – 4 (4) (-2) Disc = 49 + 32 Disc = 81

Question # 2 Find the nature of the Roots of the following given quadratic equations and the result by solving the equation.

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16x2 – 24x + 9 = 0

(v) Solution

16x2 – 24x + 9 = 0 Here, a = 16, b = -24 , c = 9 Disc = b2 – 4ac Disc = ( -24)2 – 4 (16) (9) Disc = 576 - 576 Disc = 0 The Roots are real , rational and equal Verification By quadratic formula x = x= x= x= x= x= x= So , the Roots are real rational and equal

(vi)

3x2 + 7x - 13 = 0

Solution

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3x2 + 7x - 13 = 0 Here, a = 3, b = 7 , c = -13 Disc = b2 – 4ac Disc = ( 7)2 – 4 (3) (-13) Disc = 49 + 156 Disc = 205 > 0 The Roots are real , irrational and unequal Verification By quadratic formula x = x= x= x= So , the Roots are real, irrational and unequal.

Question # 4 Find the value of k , if the roots of the following equations are equal (i)

(2k – 1)x2 + 3kx + 3 = 0

Solution (2k – 1)x2 + 3kx + 3 = 0 Here,

a = 2k -1 , b = 3k , c = 3

Given that roots are equal , so Disc = 0 b2 – 4ac = 0

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( 3k)2 – 4 (2k -1) (3) = 0 9k2 – 12(2k -1)= 0 9k2 – 24k +12= 0 3(3k2 – 8k + 4) = 0 3k2 – 8k + 4 = 0/4 3k2 – 8k + 4 = 0 3k2 – 2k – 6k + 4 = 0 K(3k -2) -2 ( 3k – 2) = 0 ( 3k – 2) (k – 2) = 0

( ii)

3k – 2= 0

0r

k–2=0

3k = 0 + 2

0r

k=0+2

3k = 2

or

k=2

K = 2/3

or

k=2

x2 +2( k + 2)x + 3k + 4 = 0

Solution x2 +2( k + 2)x + 3k + 4 = 0 Here,

a = 1 , b = 2(k +2), c = 3k + 4

Given that roots are equal , so Disc = 0 b2 – 4ac = 0 [2( k + 2)]2 – 4 (1) (3k + 4) = 0 4[ (k)2 + 2(k)(2) + (2)2] – 12k – 16 = 0 4k2 + 16k + 16 – 12k – 16 = 0 4k2 + 4k = 0 Mudassar Nazar Notes Published by Asghar Ali

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4k( k + 1) = 0 4k = 0

0r

k +1 = 0

k = 0/4

0r

k = 0 -1

k=0 ( iii)

or

k = -1

(3k + 2)x2 -5( k + 1)x + (2k + 3) = 0

Solution (3k + 2)x2 -5( k + 1)x + (2k + 3) = 0 Here, a = (3k + 2) , b = -5( k + 1) , c = 2k + 3 Given that roots are equal , so Disc = 0 b2 – 4ac = 0 [-5( k + 1)]2 – 4 (3k + 2) (2k + 3) = 0 25(k + 1)2 -4[ 3k ( 2k + 3) + 2 ( 2k + 3)] = 0 25[(k)2 + 2(k)(1) + (1)2] – 4[ 6k2 + 9k + 4k + 6] = 0 25(k2 + 2k + 1) – 4(6k2 +13k + 6) = 0 25k2 + 50k + 25 – 24k2 -52k -24 = 0 k2 - 2k + 1 = 0 (k - 1)2 = 0 = K -1 = 0 K=0+1 K=1 Question # 8 Show that the roots of the following equations are rational. (i)

a(b – c)x2 + b(c –a)x + c(a – b) = 0

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Solution a(b – c)x2 + b(c –a)x + c(a – b) = 0 The roots will be rational, if Disc is a perfect square Disc = [b(c-a)]2 – 4a(b-c) c(a-b) Disc = b2(c –a)2 – 4ac( b –c) ( a – b) Disc = b2[(c)2 - 2(c)(a) + a2 ] – 4ac[ab – b2 –ca + bc] Disc = b2c2 - 2ab2c+ a2b2 – 4a2bc + 4ab2c + 4a2c2 – 4 abc2 Disc = b2c2 + 2ab2c + a2b2 – 4a2bc - 4abc2 + 4a2c2 Disc =(a2b2 + 2ab2c +b2c2) – 4ac(ab + bc) + 4a2c2 Disc = (ab + bc)2 – 2 (2ac)(ab + bc) + (2ac)2 Disc = (ab + bc)2 – 2(ab + bc) (2ac) + (2ac)2 Disc = [(ab + bc) – 2ac]2

(Perfect square)

Hence, the roots are rational. (a + 2b)x2 + 2(a + b + c)x + (a + 2c) = 0

(ii) Solution

(a + 2b)x2 + 2(a + b + c)x + (a + 2c) = 0 The roots will be rational, if Disc is a perfect square Disc = [2(a + b + c)]2 – 4(a + 2b)(a+ 2c) Disc = 4[ a2 + b2 + c2 + 2ab + 2bc + 2ca ] – 4[a2 + 2ac + 2ab + 4bc ] Disc = 4[ a2 + b2 + c2 + 2ab + 2bc + 2ac - a2 + 2ac + 2ab + 4bc ] Disc = 4[ b2 + c2 – 2bc ] Disc = [2(b – c)]2 is a perfect square Hence, The roots are real and rational

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