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 $\left\{\begin{array}{r}\because 30=-5 \times(-6) \\ -11=-5-6\end{array}\right\}$
⇒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 :
⇒$x^{2}=\frac{25}{4}$
∴$x=\pm \sqrt{\frac{25}{4}}=\pm \frac{5}{2}$
∴$x=\frac{5}{2}, \frac{-5}{2}$

Question 2

(i) 2x2-5x=0
Sol :
⇒2x2-5x=0
⇒x(2x-5)=0
Either ,x=0
or 2x-5=0, then 2x=5
⇒$x=\frac{5}{2}$
∴$x=0, \frac{5}{2}$

(ii) x2-2x=48
Sol :
⇒x2-2x-48=0
⇒x2-8x+6x-48=0 $\left\{\begin{array}{r}\because-48=-8 \times 6 \\ -2=-8+6\end{array}\right\}$
⇒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
⇒$x=\frac{1}{2}$
∴$x=1, \frac{1}{2}$

Question 4

(i) 3x2-=2x+8
Sol :
⇒3x2=2x+8
⇒3x2-6x+4x-8=0 $\left\{\begin{array}{c}\because-8 \times 3=-24 \\ -24=-6 \times 4 \\ -2=-6+4\end{array}\right\}$
⇒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

⇒$x=\frac{-4}{3}$
∴$x=2, \frac{-4}{3}$

(ii) 4x2+15=16x
⇒4x2-16x+15=0
⇒4x2-6x-10x+15=0 $\left\{\begin{aligned} \because 4 \times 15 &=60 \\-16 &=-6+(-10) \\-16 &=-6-10 \end{aligned}\right\}$
⇒2x(2x-3)-5(2x-3)=0
⇒(2x-3)(2x-5)=0

Either , 2x-3=0, then 2x=3
⇒$x=\frac{3}{2}$
or 2x-5=0, then 2x=5

⇒$x=\frac{5}{2}$
∴$x=\frac{3}{2}, \frac{5}{2}$

Question 5

(i) x(2x+5)=25
Sol :
⇒2x2+5x-25=0
⇒2x2+10x-5x-25=0 $\left\{\begin{array}{c}\because-25 \times 2=-50 \\ -50=10 \times(-5) \\ 5=10-5\end{array}\right\}$
⇒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

⇒$x=\frac{5}{2}$
∴$x=-5, \frac{5}{2}$

(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 $\left\{\begin{array}{c}\because-18 \times 2=-36 \\ \therefore-36=-9 \times 4 \\ -5=-9+4\end{array}\right\}$
⇒x(2x-9)+2(2x-9)=0
⇒(2x-9)(x+2)=0

Either, 2x-9=0, then 2x=9
⇒$x=\frac{9}{2}$
or x+2=0, then x=-2

∴$x=\frac{9}{2},-2$

(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
⇒$x=\frac{-11}{6}$

∴$x=0,\frac{-11}{6}$

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 $x=\frac{-1}{2}$
∴$x=\frac{-1}{2}, \frac{-1}{2}$

(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 $\left\{\begin{array}{c}\because-128=-16 \times 8 \\ -8=-16+8\end{array}\right\}$
⇒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 $\left\{\begin{array}{c}\because 21 \times(-4)=-84 \\ \therefore-84=-14 \times 6 \\ -8=-14+6\end{array}\right\}$
⇒7x(3x-2)+2(3x-2)=0
⇒(3x-2)(7x+2)=0

Either, 3x-2=0, then 3x=2
⇒$x=\frac{2}{3}$

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

⇒$x=\frac{-2}{7}$

∴$x=\frac{2}{3}, \frac{-2}{7}$


(ii) $\frac{2}{3}x^2-\frac{1}{3} x-1=0$
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
⇒$x=\frac{3}{2}$
or x+1=0, then x=-1

∴$x=\frac{3}{2},-1$

Question 9

(i) $6 x+29=\frac{5}{x}$
Sol :
⇒6x2+29x-5=0
⇒6x2+30x-x-5=0 $\left\{\begin{array}{c}\because 6 \times(-5)=-30 \\ \therefore-30=30 \times(-1) \\ 29=30-1\end{array}\right\}$
⇒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 
⇒$x=\frac{1}{6}$

∴$x=\frac{1}{5},-5$


(ii) $x+\frac{1}{x}=2 \frac{1}{2}$
Sol :
⇒$x+\frac{1}{x}=\frac{5}{2}$
⇒$x^{2}+1=\frac{5}{2} x$
⇒$x^{2}-\frac{5}{2} x+1=0$
⇒2x2-5x+2=0
⇒2x2-x-4x+2=0 $\left\{\begin{array}{c}\because 2 \times 2=4 \\ 4=-1 \times(-4) \\ -5=-1-4\end{array}\right\}$
⇒x(2x-1)-2(2x-1)=0
⇒(2x-1)(x-2)=0

Either ,2x-1=0, then 2x=1
⇒$x=\frac{1}{2}$ or x-2=0, then x=2

∴$x=2, \frac{1}{2}$

Question 10

(i) $3 x-\frac{8}{x}=2$
Sol :
⇒$\frac{3 x^{2}-8=2 x}{x}$
⇒3x2-2x-8=0
⇒3x2-6x+4x-8=0 $\left\{\begin{array}{c}\because-8 \times 3=-24 \\ -24=-6 \times 4 \\ -2=-6+4\end{array}\right\}$
⇒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
⇒$x=\frac{-4}{3}$

∴$x=2, \frac{-4}{3}$


(ii) $\frac{x}{3}+\frac{9}{x}=4$
Sol :
⇒$\frac{x^{2}+27=12 x}{3 x}$
⇒x2-12x+27=0
⇒x2-3x-9x+27=0 $\left\{\begin{array}{r}\because 27=-3 \times(-9) \\ -12=-3-9\end{array}\right\}$ 
⇒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) $\frac{x-1}{x+1}=\frac{2 x-5}{3 x-7}$
Sol :
⇒$\frac{x-1}{x+1}=\frac{2 x-5}{3 x-7}$
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 $\left\{\begin{array}{c}\because 12=-4 \times(-3) \\ -7=-4-3\end{array}\right\}$
⇒(x-4)(x-3)=0

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

∴x=3 ,4


(ii) $\frac{1}{x+2}+\frac{1}{x}=\frac{3}{4}$
Sol :
⇒$\frac{x+x+2}{x(x+2)}=\frac{3}{4}$
⇒$\frac{2 x+2}{x(x+2)}=\frac{3}{4}$

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 $\left\{\begin{array}{c}\because 3 \times(-8)=-24 \\ \therefore-24=-6 \times 4 \\ -2=-6+4\end{array}\right\}$

⇒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
⇒$x=\frac{-4}{3}$

Hence , $x=2, \frac{-4}{3}$

Question 12

(i) $\frac{8}{x+3}-\frac{3}{2-x}=2$
Sol :
⇒$\frac{8(2-x)-3(x+3)}{(x+3)(2-x)}=\frac{2}{1}$
⇒$\frac{16-8 x-3 x-9}{2 x-x^{2}+6-3 x}=\frac{2}{1}$
⇒$\frac{7-11 x}{-x^{2}-x+6}=\frac{2}{1}$
⇒7-11x=-2x2-2x+12 (by cross multiplication )

⇒2x2+2x-12+7-11x=0
⇒2x2-9x-5=0
⇒2x2-10x+x-5=0 $\left\{\begin{array}{c}\because 2 \times(-5)=-10 \\ \therefore-10=-10 \times 1 \\ -9=-10+1\end{array}\right\}$
⇒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

⇒$x=\frac{-1}{2}$

∴$x=5, \frac{-1}{2}$


(ii) $\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{1}{6}$
Sol :
⇒$\frac{x^{2}+(x+1)^{2}}{x(x+1)}=\frac{13}{6}$
⇒$\frac{x^{2}+x^{2}+2 x+1}{x^{2}+x}=\frac{13}{6}$
⇒$\frac{2 x^{2}+2 x+1}{x^{2}+x}=\frac{13}{6}$
⇒13x2+13x=12x2+12x+6
⇒13x2+13x-12xx2-12x-6=0
⇒x2+x-6=0
⇒x2+3x-2x-6=0 $\left\{\begin{array}{c}\because-6=3 \times(-2) \\ 1=3-2\end{array}\right\}$
⇒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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