ML Aggarwal Solution Class 9 Chapter 7 Quadratic Equations Exercise 7.1

 Exercise 7.1

Question 1

Solve the following (1 to 12) equations:

(i) x2– 11x + 30 = 0
Sol :
⇒x2-11x+30=0
⇒x2-5x-6x+30=0 {30=5×(6)11=56}
⇒x(x-5)-6(x-5)=0
⇒(x-5)(x-6)=0
Either ,x-5=0, then x=5
or x-6=0, then x=6
∴x=5 , 6
 
(ii) 4x2-25 = 0
Sol :
x2=254
x=±254=±52
x=52,52

Question 2

(i) 2x2-5x=0
Sol :
⇒2x2-5x=0
⇒x(2x-5)=0
Either ,x=0
or 2x-5=0, then 2x=5
x=52
x=0,52

(ii) x2-2x=48
Sol :
⇒x2-2x-48=0
⇒x2-8x+6x-48=0 {48=8×62=8+6}
⇒x(x-8)+6(x-8)=0
⇒(x-8)(x+6)=0
Either ,x-8=0, then x=8
or x+6=0, then x=-6
∴x=8, -6

Question 3

(i) 6+x=x2
Sol :
⇒6+x=x2
⇒x2-x-6=0
⇒x2-3x+2x-6=0
⇒x(x-3)+2(x-3)=0
⇒(x-3)(x+2)=0
Either ,x-3=0, then x=3
or x+2=0, then x=-2
∴x=3, 2

(ii) 2x2+3x+1=0
Sol :
⇒2x2-2x-x+1=0
⇒2x(x-1)-1(x-1)=0
⇒(x-1)(2x-1)=0
Either, x-1=0, then x=1
or 2x-1=0, then 2x=1
x=12
x=1,12

Question 4

(i) 3x2-=2x+8
Sol :
⇒3x2=2x+8
⇒3x2-6x+4x-8=0 {8×3=2424=6×42=6+4}
⇒3x(x-2)+4(x-2)=0
⇒(x-2)(3x+4)=0

Either, x-2=0, then x=2
or 3x+4=0,$ then 3x=-4

x=43
x=2,43

(ii) 4x2+15=16x
⇒4x2-16x+15=0
⇒4x2-6x-10x+15=0 {4×15=6016=6+(10)16=610}
⇒2x(2x-3)-5(2x-3)=0
⇒(2x-3)(2x-5)=0

Either , 2x-3=0, then 2x=3
x=32
or 2x-5=0, then 2x=5

x=52
x=32,52

Question 5

(i) x(2x+5)=25
Sol :
⇒2x2+5x-25=0
⇒2x2+10x-5x-25=0 {25×2=5050=10×(5)5=105}
⇒2x(x+5)-5(x+5)=0
⇒(x+5)(2x-5)=0

Either ,x+5=0, then x=-5
or 2x-5=0, then 2x=5

x=52
x=5,52

(ii) (x+3) (x–3)=40
Sol :
⇒x2-9=40
⇒x2-9-40=0
⇒x2-49=0
⇒(x)2-(7)2=0

Either , x+7=0, then x=-7
or x-7=0, then x=7
∴x=7,-7

Question 6

(i) (2x + 3) (x – 4) = 6
Sol :
⇒(2 x+3)(x-4)=6
⇒2x2-8 x+3 x-12-6=0
⇒2x2-5 x-18=0
⇒2x2-9 x+4 x-18=0 {18×2=3636=9×45=9+4}
⇒x(2x-9)+2(2x-9)=0
⇒(2x-9)(x+2)=0

Either, 2x-9=0, then 2x=9
x=92
or x+2=0, then x=-2

x=92,2

(ii) (3x + 1) (2x + 3) = 3
Sol :
⇒6x2+9x+2x+3-3=0
⇒6x2+11x=0
⇒x(6x+11)=0
Either, x=0
or 6x+11=0, then 6x=-11
x=116

x=0,116

Question 7

(i) 4x2+4x+1=0
Sol :
⇒4x2+2x+2x+1=0
⇒2x(2x+1)+1(2x+1)=0
⇒(2x+1)(2x+1)=0
Either , 2x+1=0, then x=12
x=12,12

(ii) (x-4)2+52=132
Sol :
⇒x2-8x+16+25=169
⇒x2-8x+16+25-169=0
⇒x2-8x-128=0
⇒x2-16x+8x-128=0 {128=16×88=16+8}
⇒x(x-16)+8(x-16)=0
⇒(x-16)(x+8)=0

Either ,x-16=0, then x=16
or x+8=0, then x=-8

∴x=16, 8

Question 8

(i) 21x2=4(2x+1)
Sol :
⇒21x2=4(2x+1)
⇒21x2-8x-4=0
⇒21x2-14x+6x-4=0 {21×(4)=8484=14×68=14+6}
⇒7x(3x-2)+2(3x-2)=0
⇒(3x-2)(7x+2)=0

Either, 3x-2=0, then 3x=2
x=23

or 7x+2=0, then 7x=-2

x=27

x=23,27


(ii) 23x213x1=0
Sol :
⇒2x2-x-3=0
⇒2x2-3x+2x-3=0
⇒x(2x-3)+1(2x-3)=0
⇒(2x-3)(x+1)=0

Either,2x-3=0, then 2x=3
x=32
or x+1=0, then x=-1

x=32,1

Question 9

(i) 6x+29=5x
Sol :
⇒6x2+29x-5=0
⇒6x2+30x-x-5=0 {6×(5)=3030=30×(1)29=301}
⇒6x(x+5)-1(x+5)=0
⇒(x+5)(6x-1)=0

Either,x=+5=0, then x=-5
or 6x-1=0, then 6x=1 
x=16

x=15,5


(ii) x+1x=212
Sol :
x+1x=52
x2+1=52x
x252x+1=0
⇒2x2-5x+2=0
⇒2x2-x-4x+2=0 {2×2=44=1×(4)5=14}
⇒x(2x-1)-2(2x-1)=0
⇒(2x-1)(x-2)=0

Either ,2x-1=0, then 2x=1
x=12 or x-2=0, then x=2

x=2,12

Question 10

(i) 3x8x=2
Sol :
3x28=2xx
⇒3x2-2x-8=0
⇒3x2-6x+4x-8=0 {8×3=2424=6×42=6+4}
⇒3x(x-2)+4(x-2)=0
⇒(x-2)(3x+4)=0

Either, x-2=0, then x=2
or 3x+4=0, then 3x=-4
x=43

x=2,43


(ii) x3+9x=4
Sol :
x2+27=12x3x
⇒x2-12x+27=0
⇒x2-3x-9x+27=0 {27=3×(9)12=39} 
⇒x(x-3)-9(x-3)=0
⇒(x-3)(x-9)=0

Either ,x-3=0, then x=3
or x-9=0, then x=9

Question 11

(i) x1x+1=2x53x7
Sol :
x1x+1=2x53x7
By cross multiplication
⇒(x-1)(3x-7)=(x+1)(2x-5)
⇒3x2-7x-3x+7=2x2-5x+2x-5
⇒3x2-10x+7=2x2-3x-5=0
⇒3x2-10x+7-2x2+3x+5=0
⇒x2-7x+12=0
⇒x2-4x-3x+12=0
⇒x(x-4)-3(x-4)=0 {12=4×(3)7=43}
⇒(x-4)(x-3)=0

Either ,x--4=0, then x=4
or x-3=0, then x=3

∴x=3 ,4


(ii) 1x+2+1x=34
Sol :
x+x+2x(x+2)=34
2x+2x(x+2)=34

by cross multiplication
⇒3x(x+2)=4(2x+2)
⇒3x2+6x=8x+8
⇒3x2+6x-8x-8=0
⇒3x2-2x-8=0
⇒3x2-6x+4x-8=0 {3×(8)=2424=6×42=6+4}

⇒3x(x-2)+4(x-2)=0
⇒(x-2)(3x+4)=0

Either , x-2=0, then x=2
or 3x+4=0, then 3x=-4
x=43

Hence , x=2,43

Question 12

(i) 8x+332x=2
Sol :
8(2x)3(x+3)(x+3)(2x)=21
168x3x92xx2+63x=21
711xx2x+6=21
⇒7-11x=-2x2-2x+12 (by cross multiplication )

⇒2x2+2x-12+7-11x=0
⇒2x2-9x-5=0
⇒2x2-10x+x-5=0 {2×(5)=1010=10×19=10+1}
⇒2x(x-5)+1(x-5)=0
⇒(x-5)(2x+1)=0

Either , x-5=0, then x=5
or 2x+1=0, then 2x=-1

x=12

x=5,12


(ii) xx+1+x+1x=216
Sol :
x2+(x+1)2x(x+1)=136
x2+x2+2x+1x2+x=136
2x2+2x+1x2+x=136
⇒13x2+13x=12x2+12x+6
⇒13x2+13x-12xx2-12x-6=0
⇒x2+x-6=0
⇒x2+3x-2x-6=0 {6=3×(2)1=32}
⇒x(x+3)-2(x+3)=0
⇒(x+3)(x-0)=0
Either ,x+3=0, then x=-3
or x-2=0, then x=2

∴x=2,3

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