Solve each of the following equations:

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

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

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

$\displaystyle 4x^2+6x+x-3+3x+9=0$

$\displaystyle 4x^2+10x+6=0$

$\displaystyle 4x^2+6x+4x+6=0$

$\displaystyle 4x(x+1)+6(x+1)=0$

$\displaystyle (4x+6)((x+1)=0$

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

$\displaystyle \\$

Question 2: $\displaystyle (2x+3)^2=81$

$\displaystyle (2x+3)^2=81$

$\displaystyle 4x^2+9+12x=81$

$\displaystyle 4x^2+12x-72=0$

$\displaystyle x^2+3x-18=0$

$\displaystyle x^2+6x-3x-18=0$

$\displaystyle x(x+6)-3(x+6)=0$

$\displaystyle (x-3)(x+6)=0$

$\displaystyle \Rightarrow x = 3, -6$

$\displaystyle \\$

Question 3: $\displaystyle a^2x^2-b^2=0$

$\displaystyle a^2x^2-b^2=0$

$\displaystyle (ax-b)(ax+b)=0$

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

$\displaystyle \\$

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

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

$\displaystyle 8x^2-22x+15=0$

$\displaystyle 8x^2-10x-12x+15=0$

$\displaystyle 4x(2x-3)-5(2x-3)=0$

$\displaystyle (2x-3)(4x-5)=0$

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

$\displaystyle \\$

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

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

$\displaystyle x^2+4x+4=0$

$\displaystyle x^2+2x+2x+4=0$

$\displaystyle (x+2)(x+2)=0$

$\displaystyle \Rightarrow x = -2$

$\displaystyle \\$

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

$\displaystyle 2x^4-5x^2+3 = 0$

Let $\displaystyle x^2=y$

$\displaystyle 2y^2-5y+3=0$

$\displaystyle 2y^2-3y-2y+3=0$

$\displaystyle 2y(y-1)-3(y-1)=0$

$\displaystyle (2y-3)(y-1)=0$

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

$\displaystyle \text{When } y = 1$

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

$\displaystyle \text{When } y = \frac{3}{2}$

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

$\displaystyle \\$

Question 7: $\displaystyle x^4-2x^2-3=0$

$\displaystyle x^4-2x^2-3=0$

Let $\displaystyle x^2 = y$

$\displaystyle y^2-2y-3=0$

$\displaystyle y^2-3y+y-3=0$

$\displaystyle y(y-3)+(y-3)=0$

$\displaystyle (y+1)(y-3)=0$

$\displaystyle \Rightarrow y = -1, 3$

$\displaystyle \text{When } y = -1$ , $\displaystyle x$ is imaginary

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \text{Therefore } 9(y^2-2)-9y-52=0$

$\displaystyle 9y^2-9y-70=0$

$\displaystyle 9y^2-30y+21y-70=0$

$\displaystyle 3y(3y-10)+7(3y-10)=0$

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

$\displaystyle \text{When } y = \frac{10}{3}$

$\displaystyle x+\frac{1}{x} = \frac{10}{3}$

$\displaystyle 3x^2-10x+3=0$

$\displaystyle 3x^2-9x-x+3=0$

$\displaystyle 3x(x-3)-(x-3)=0$

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

$\displaystyle \text{When } y =-\frac{7}{3}$

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

$\displaystyle 3x^2+7x+3=0$

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

$\displaystyle \\$

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

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

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

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

$\displaystyle 2(y^2-2)-y=11$

$\displaystyle 2y^2-y-15=0$

$\displaystyle 2y^2-6y+5y-15=0$

$\displaystyle 2y(y-3)+5(y-3)=0$

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

$\displaystyle \text{When } y = 3$

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

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

$\displaystyle \text{When } y = -\frac{5}{2}$

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

$\displaystyle 2x^2+5x+2=0$

$\displaystyle 2x^2+4x+x+2=0$

$\displaystyle 2x(x+2)+(x+2)=0$

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

$\displaystyle \\$

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

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

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

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

$\displaystyle (y^2+2)-3y-2=0$

$\displaystyle y^2-3y=0$

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

$\displaystyle \text{When } y = 0$

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

$\displaystyle x^2-1=0 \Rightarrow x = \pm 1$

$\displaystyle \text{When } y =3$

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

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

$\displaystyle \\$

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

$\displaystyle (x^2+5x+4)(x^2+5x+6)=120$

Let $\displaystyle x^2+5x=y$

$\displaystyle (y+4)(y+6)=120$

$\displaystyle y^2+10y-96=0$

$\displaystyle y^2-6y+16y-96=0$

$\displaystyle y(y-6)+16(y-6)=0$

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

$\displaystyle \text{When } y = 6$

$\displaystyle x^2+5x-6=0$

$\displaystyle x^2+6x-x-6=0$

$\displaystyle x(x+6)-(x+6)=0$

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

$\displaystyle \text{When } y = -16$

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

$\displaystyle \\$

Question 12: $\displaystyle x^2-5x-10=0$ [2005]

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

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

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

Solving we get $\displaystyle x = 6.53, -1.53$

$\displaystyle \\$

Question 13: $\displaystyle 3x^2-x-7=0$ [2004]

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

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

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

Solving we get $\displaystyle x = 1.703, -1.3699$

$\displaystyle \\$

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

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

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

$\displaystyle y^2-7y+12=0$

$\displaystyle y^2-3y-4y+12=0$

$\displaystyle y(y-3)-4(y-3)=0$

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

$\displaystyle \text{When } y = 3$

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

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

$\displaystyle \text{When } y =4$

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

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

$\displaystyle \\$

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

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

$\displaystyle x^2-12x+x-12=0$

$\displaystyle x(x-12)+(x-12)=0$

$\displaystyle (x-12)(x+1)=0$

$\displaystyle x=12, -1$ $\displaystyle \text{Since } x \in N, x = 12$

$\displaystyle \\$

Question 16: $\displaystyle x^2-4x-12=0; \text{ when } x \in I$

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

$\displaystyle x^2-6x+2x-12=0$

$\displaystyle x(x-6)+2(x-6)=0$

$\displaystyle (x-6)(x+2)=0 \Rightarrow x = 6, -2$

$\displaystyle \\$

Question 17: $\displaystyle 2x^2-9x+10=0; \text{ when } x \in Q$

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

$\displaystyle 2x^2-4x-5x+10=0$

$\displaystyle 2x(x-2)-5(x-2)=0$

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

$\displaystyle \\$

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

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

Let $\displaystyle (a+b)x = y$

$\displaystyle y^2-y-6=0$

$\displaystyle y^2-3y+2y-6=0$

$\displaystyle y(y-3)+2(y-3)=0$

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

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle x(x+p+q)+pq=0$

$\displaystyle x^2+(p+q)x+pq = 0$

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

$\displaystyle \\$

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

$\displaystyle x(x+1)+(x+2)(x+3)=42$

$\displaystyle x^2+x+x^2+5x+6 = 42$

$\displaystyle 2x^2+6x-36=0$

$\displaystyle x^2+3x-18=0$

$\displaystyle x^2-3x+6x-18=0$

$\displaystyle x(x-3)+6(x-3)=0$

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

$\displaystyle \\$

$\displaystyle \text{Question 1: } \frac{1}{x+1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}$

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

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

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

$\displaystyle (x+3)(x+4)=(x+1)(x+2)$

$\displaystyle x^2+7x+12=x^2+3x+2$

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

$\displaystyle \\$

Find $\displaystyle m \text{ or } p$ so that the equation have equal roots:

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

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

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

$\displaystyle (-4)^2-4(m-3)(1)=0$

$\displaystyle 16-4m+12=0 \Rightarrow 7$

Solution: $\displaystyle 4x^2-4x+1=0$

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

$\displaystyle \\$

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

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

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

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

$\displaystyle 144-12m-84=0$

$\displaystyle 12m = 60 \Rightarrow m = 5$

Solving the equation

$\displaystyle 3x^2+12x+12=0$

$\displaystyle x^2+4x+4=0$

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

$\displaystyle \\$

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

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

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

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

$\displaystyle m^2+4+4m-4m-20=0$

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

$\displaystyle \text{When } m=4$

$\displaystyle x^2-6x+9=0$

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

$\displaystyle \text{When } m = -4$

$\displaystyle x^2+2x+1=0$

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

$\displaystyle \\$

Question 25: $\displaystyle px^2-4x+3=0$ [2010]

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

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

$\displaystyle (-4)^2-4(p)(3)=0$

$\displaystyle 16-12p=0$

$\displaystyle p=\frac{4}{3}$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle 4m^2+4-8m-4m-20=0$

$\displaystyle 4m^2-12m-16=0$

$\displaystyle m^2-3m-4=0$

$\displaystyle m^2-4m+m-4=0$

$\displaystyle m(m-4)+(m-4)=0$

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