Click Here (Amazon): Books for ICSE Class 10 Board Exams

Solve each of the following equations:

Question 1: \frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0   where x \neq 3 \ and \  x \neq -\frac{3}{2}

Answer:

\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0   where x \neq 3 \ and \  x \neq -\frac{3}{2}

\frac{2x(2x+3)+(x-3)}{(x-3)(2x+3)} + \frac{3x+9}{(x-3)(2x+3)}=0

4x^2+6x+x-3+3x+9=0

4x^2+10x+6=0

4x^2+6x+4x+6=0

4x(x+1)+6(x+1)=0

(4x+6)((x+1)=0 

\Rightarrow x = -\frac{3}{2}, -1

\\

Question 2: (2x+3)^2=81

Answer:

(2x+3)^2=81

4x^2+9+12x=81

4x^2+12x-72=0

x^2+3x-18=0

x^2+6x-3x-18=0

x(x+6)-3(x+6)=0

(x-3)(x+6)=0

\Rightarrow x = 3, -6

\\

Question 3: a^2x^2-b^2=0

Answer:

a^2x^2-b^2=0

(ax-b)(ax+b)=0

\Rightarrow x = \frac{b}{a}, -\frac{b}{a}

\\

Question 4: x^2-\frac{11}{4}x+\frac{15}{8}=0

Answer:

x^2-\frac{11}{4}x+\frac{15}{8}=0  Multiplying the equation by 8.

8x^2-22x+15=0

8x^2-10x-12x+15=0

4x(2x-3)-5(2x-3)=0

(2x-3)(4x-5)=0

\Rightarrow x = \frac{3}{2}, \frac{5}{4}

\\

Question 5: x+\frac{4}{x}=-4; x \neq 0

Answer:

x+\frac{4}{x}=-4; x \neq 0

x^2+4x+4=0

x^2+2x+2x+4=0

(x+2)(x+2)=0

\Rightarrow x = -2

\\

Question 6: 2x^4-5x^2+3 = 0

Answer:

2x^4-5x^2+3 = 0

Let x^2=y

2y^2-5y+3=0

2y^2-3y-2y+3=0

2y(y-1)-3(y-1)=0

(2y-3)(y-1)=0

\Rightarrow y = \frac{3}{2}, 1

When y = 1

x^2=1 \Rightarrow x =\pm 1

When y = \frac{3}{2}

x^2=\frac{3}{2} \Rightarrow x = \pm \sqrt{\frac{3}{2}}

\\

Question 7: x^4-2x^2-3=0

Answer:

x^4-2x^2-3=0

Let x^2 = y

y^2-2y-3=0

y^2-3y+y-3=0

y(y-3)+(y-3)=0

(y+1)(y-3)=0

\Rightarrow y = -1, 3

When y = -1 , x is imaginary

When y = 3 , then x= \pm \sqrt{3}

\\

Question 8: 9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0

Answer:

9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0

Let x + \frac{1}{x}=y

x^2+\frac{1}{x^2}+2=y^2

x^2+\frac{1}{x^2}=y^2-2

Therefore 9(y^2-2)-9y-52=0

9y^2-9y-70=0

9y^2-30y+21y-70=0

3y(3y-10)+7(3y-10)=0

(3y-10)(3y+7)=0 \Rightarrow y = \frac{10}{3}, -\frac{7}{3}

When y = \frac{10}{3}

x+\frac{1}{x} = \frac{10}{3}

3x^2-10x+3=0

3x^2-9x-x+3=0

3x(x-3)-(x-3)=0

(3x-1)(x-3)=0 \Rightarrow x = \frac{1}{3}, 3

When y =-\frac{7}{3}

x+\frac{1}{x}=-\frac{7}{3}

3x^2+7x+3=0

Therefore x = \frac{-7 \pm \sqrt{13}}{6}

\\

Question 9: 2(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=11

Answer:

2(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=11

y = x+\frac{1}{x}

y^2 = x^2+\frac{1}{x^2}+2

2(y^2-2)-y=11

2y^2-y-15=0

2y^2-6y+5y-15=0

2y(y-3)+5(y-3)=0

(y-3)(2y+5)=0 \Rightarrow y = 3, -\frac{5}{2}

When y = 3

x+\frac{1}{x}=3

x^2-3x+1=0 \Rightarrow x = 3 \pm \frac{\sqrt{5}}{2}

When y = -\frac{5}{2}

x+\frac{1}{x}=-\frac{5}{2}

2x^2+5x+2=0

2x^2+4x+x+2=0

2x(x+2)+(x+2)=0

(x+2)(2x+1)=0 \Rightarrow x = -2, -\frac{1}{2}

\\

Question 10: (x^2+\frac{1}{x^2})-3(x+\frac{1}{x})-2=0

Answer:

(x^2+\frac{1}{x^2})-3(x+\frac{1}{x})-2=0

Let x -\frac{1}{x}=y

x^2+\frac{1}{x^2} = y^2+2

(y^2+2)-3y-2=0

y^2-3y=0

y(y-3)=0 \Rightarrow y = 0, 3

When y = 0

x-\frac{1}{x}=0

x^2-1=0 \Rightarrow  x = \pm 1

When y =3

x-\frac{1}{x}=3

x^2-3x-1=0 \Rightarrow \frac{3 \pm \sqrt{5}}{2}

\\

Question 11: (x^2+5x+4)(x^2+5x+6)=120

Answer:

(x^2+5x+4)(x^2+5x+6)=120

Let x^2+5x=y

(y+4)(y+6)=120

y^2+10y-96=0

y^2-6y+16y-96=0

y(y-6)+16(y-6)=0

(y-6)(y+16)=0 \Rightarrow y =6, -16

When y = 6

x^2+5x-6=0

x^2+6x-x-6=0

x(x+6)-(x+6)=0

(x+6)(x-1)=0 \Rightarrow x = -6, 1

When y = -16

x^2+5x+16=0 \Rightarrow x is imaginary

\\

Question 12: x^2-5x-10=0     [2005]

Answer:

Comparing x^2-5x-10=0 with ax^2+bx+c=0 , we get a = 1, b = -5 \ and \ c =-10

Since  x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Therefore  x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}

Solving we get x = 6.53, -1.53

\\

Question 13: 3x^2-x-7=0     [2004]

Answer:

Comparing 3x^2-x-7=0 with ax^2+bx+c=0 , we get a = 3, b = -1 \ and \ c =-7

Since  x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Therefore  x = \frac{-(-1) \pm \sqrt{(-1)^2-4(3)(-7)}}{2(3)}

Solving we get x = 1.703, -1.3699

\\

Question 14: (\frac{x}{x+2})^2-7(\frac{x}{x+2})+12=0, x \neq -2

Answer:

(\frac{x}{x+2})^2-7(\frac{x}{x+2})+12=0, x \neq -2

Let \frac{x}{x+2}=y

y^2-7y+12=0

y^2-3y-4y+12=0

y(y-3)-4(y-3)=0

(y-3)(y-4)=0 \Rightarrow y = 3, 4

When y = 3

\frac{x}{x+2}=3

3x+6=x \Rightarrow x = -3

When y =4

\frac{x}{x+2}=4

4x+8=x \Rightarrow x = -\frac{8}{3}

\\

Question 15: x^2-11x-12=0; \ when \ x \in N

Answer:

x^2-11x-12=0; \ when \ x \in N

x^2-12x+x-12=0

x(x-12)+(x-12)=0

(x-12)(x+1)=0

x=12, -1 Since x \in N, x = 12 

\\

Question 16: x^2-4x-12=0; \ when \ x \in I

Answer:

x^2-4x-12=0; \ when \ x \in I

x^2-6x+2x-12=0

x(x-6)+2(x-6)=0

(x-6)(x+2)=0 \Rightarrow x = 6, -2

\\

Question 17: 2x^2-9x+10=0; \ when \ x \in Q

Answer:

2x^2-9x+10=0; \ when \ x \in Q

2x^2-4x-5x+10=0

2x(x-2)-5(x-2)=0

(x-2)(2x-5)=0 \Rightarrow x = 2, \frac{5}{2}

\\

Question 18: (a+b)^2x^2-(a+b)x-6=0; a+b \neq 0

Answer:

(a+b)^2x^2-(a+b)x-6=0; a+b \neq 0

Let (a+b)x = y

y^2-y-6=0

y^2-3y+2y-6=0

y(y-3)+2(y-3)=0

(y-3)(y+2)=0 \Rightarrow y = 3, -2

When y =3 \Rightarrow (a+b)x = 3 \Rightarrow x = \frac{3}{a+b}

When y = -2 \Rightarrow (a+b)x = -2 \Rightarrow x =-\frac{2}{a+b}

\\

Question 19: \frac{1}{p}+\frac{1}{q}+\frac{1}{x} = \frac{1}{x+p+q}

Answer:

\frac{1}{p}+\frac{1}{q}+\frac{1}{x} = \frac{1}{x+p+q}

\frac{1}{p}+\frac{1}{q}+\frac{1}{x} - \frac{1}{x+p+q} =0

\frac{p+q}{pq}+\frac{pq}{x(x+p+q)}=0

x(x+p+q)+pq=0

x^2+(p+q)x+pq = 0

(x+p)(x+q)=0 \Rightarrow x = -p, -q

\\

Question 20: x(x+1)+(x+2)(x+3)=42

Answer:

x(x+1)+(x+2)(x+3)=42

x^2+x+x^2+5x+6 = 42

2x^2+6x-36=0

x^2+3x-18=0

x^2-3x+6x-18=0

x(x-3)+6(x-3)=0

(x-3)(x+6)=0 \Rightarrow x = 3, -6

\\

Question 21: \frac{1}{x+1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}

Answer:

\frac{1}{x+1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}

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

\frac{x+2-2x-2}{(x+1)(x+2)} = \frac{3x+12-4x-12}{(x+3)(x+4)}

(x+3)(x+4)=(x+1)(x+2)

x^2+7x+12=x^2+3x+2

4x=-10 \Rightarrow x = -\frac{5}{2}

\\

Find m \ or \ p  so that the equation have equal roots:

Question 22: (m-3)x^2-4x+1=0

Answer:

Comparing (m-3)x^2-4x+1=0  with ax^2+bx+c=0 , we get a = (m-3), b = -4 \ and \ c =1

For roots to be equal, we should have b^2-4ac = 0

(-4)^2-4(m-3)(1)=0

16-4m+12=0 \Rightarrow 7

Solution: 4x^2-4x+1=0

(2x-1)(2x-1)=0 \Rightarrow x = \frac{1}{2}

\\

Question 23: 3x^2+12x+(m+7)=0

Answer:

Comparing 3x^2+12x+(m+7)=0  with ax^2+bx+c=0 , we get a = 3, b = 12 \ and \ c =(m+7)

For roots to be equal, we should have b^2-4ac = 0

(12)^2-4(3)(m+7)=0

144-12m-84=0

12m = 60 \Rightarrow m = 5

Solving the equation

3x^2+12x+12=0

x^2+4x+4=0

(x+2)(x+2)=0 \Rightarrow x = -2

\\

Question 24: x^2-(m+2)x+(m+5)=0

Answer:

Comparing x^2-(m+2)x+(m+5)=0  with ax^2+bx+c=0 , we get a = 1, b = -(m+2) \ and \ c =(m+5)

For roots to be equal, we should have b^2-4ac = 0

(m+2)^2-4(1)(m+5)=0

m^2+4+4m-4m-20=0

m^2=16 \Rightarrow m =\pm 4

When m=4

x^2-6x+9=0

(x-3)(x-3)=0 \Rightarrow x = 3

When m = -4

x^2+2x+1=0

(x+1)^2 = 0 \Rightarrow x = -1

\\

Question 25: px^2-4x+3=0     [2010]

Answer:

Comparing px^2-4x+3=0  with ax^2+bx+c=0 , we get a = p, b = -4 \ and \ c =3

For roots to be equal, we should have b^2-4ac = 0

(-4)^2-4(p)(3)=0

16-12p=0

p=\frac{4}{3}

\\

Question 26: x^2+2(m-1)x+(m+5)=0    [2012]

Answer:

Comparing x^2+2(m-1)x+(m+5)=0  with ax^2+bx+c=0 , we get a = 1, b = 2(m-1) \ and \ c =(m+5)

For roots to be equal, we should have b^2-4ac = 0

(2(m-1))^2-4(1)(m+5)=0

4(m^2+1-2m)-4(m+5)=0

4m^2+4-8m-4m-20=0

4m^2-12m-16=0

m^2-3m-4=0

m^2-4m+m-4=0

m(m-4)+(m-4)=0

(m-4)(m+1)=0 \Rightarrow m = 4, -1

Click Here (Amazon): Books for ICSE Class 10 Board Exams

Advertisements