ML Aggarwal Solution Class 9 Chapter 7 Quadratic Equations Test

Test

Solve the following (1 to 3) equations :

Question 1

(i) x(2x+ 5) = 3

Sol :

⇒2x2+5x-3=0

⇒2x2+6x-x-3=0 $\left\{\begin{array}{c}\because 2 \times(-3)=-6 \\ \therefore-6=6 \times(-1) \\ 5=6-1\end{array}\right\}$

⇒2x(x+3)-1(x+3)=0

⇒(x+3)(2x-1)=0

Either ,x+3=0, then x=-3

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

⇒$x=\frac{1}{2}$

∴$x=-3, \frac{1}{2}$


(ii) 3x2–4x–4=0

Sol :
⇒3x2-6x+2x-4=0

⇒3x(x-2)+2(x-2)=0 $\left\{\begin{array}{c}\because 3 \times(-4)=-12 \\ \therefore-12=-6 \times 2 \\ -4=-6+2\end{array}\right\}$

⇒(x-2)(3x+2)=0

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

or 3x+2=0, then 3x=-2

⇒$x=\frac{-2}{3}$


Question 2

(i) $4 x^{2}-2 x+\frac{1}{4}=0$

Sol :

⇒16x2-8x+1=0

⇒16x2-8x+1=0

⇒16x2-4x-4x+1=0

⇒4x(4x-1)-1(4x-1)=0

⇒(4x-1)(4x-1)=0

⇒(4x-1)2=0

∴$x=\frac{1}{4}, \frac{1}{4}$


(ii) 2x2+7x+6=0

Sol :

⇒2x2+4x+3x+6=0 $\left\{\begin{array}{c}\because 2 \times 6=12 \\ \therefore 12=3 \times 4 \\ 7=3+4\end{array}\right\}$

⇒2x(x+2)+3(x+2)=0

⇒(x+2)(2x+3)=0

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

or 2x+3=0, then 2x=-3

⇒$x=\frac{-3}{2}$

∴$x=-2, \frac{-3}{2}$


Question 3

(i) $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}$

Sol :

⇒$\frac{(x-1)(x-4)+(x-2)(x-3)}{(x-2)(x-4)}=\frac{10}{3}$

⇒$\frac{x^{2}-5 x+4+x^{2}-5 x+6}{x^{2}-6 x+8}=\frac{10}{3}$

⇒$\frac{2 x^{2}-10 x+10}{x^{2}-6 x+8}=\frac{10}{3}$

⇒10x2-60x+80=6x2-30x+30

⇒10x2-60x+80-6x2+30x-30=0

⇒4x2-30x+50=0

⇒2x2-15x+25=0

⇒2x2-10x-5x+25=0 $\left\{\begin{array}{l}\because 2 \times 25=50 \\ \therefore 50=-10 \times(-5) \\ -15=-10-5\end{array}\right\}$

⇒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) $\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}$

Sol :

⇒$\frac{6 x-6-2 x}{x(x-1)}=\frac{1}{x-2}$

⇒$\frac{4 x-6}{x^{2}-x}=\frac{1}{x-2}$

⇒(4x-6)(x-2)=x2-x

(by cross multiplication)

⇒4x2-8x-6x+12=x2-x

⇒4x2-14x+12-x2+x=0

⇒3x2-13x+12=0

⇒3x2-4x-9x+12=0 $\left\{\begin{array}{l}\because 3 \times 12=36 \\ \therefore 36=(-4) \times(-9) \\ -13=-4-9\end{array}\right\}$

⇒x(3x-4)-3(3x-4)=0

⇒(3x-4)(x-3)=0

Either, 3x-4=0, then 3x=4 

⇒$x=\frac{4}{3}$

or x-3=0, then x=3

∴$x=3, \frac{4}{3}$

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