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