Solve each of the following equations for \displaystyle x  

Question 1: \displaystyle x^2-8x+5=0  

Answer:

Comparing \displaystyle x^2-8x+5=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 1, b = -8 \text{ and } c =5  

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

\displaystyle \text{Therefore } x = \frac{-(-8) \pm \sqrt{(-8)^2-4(1)(5)}}{2(1)}  

\displaystyle \text{Solving we get } x = 7.32, 0.68  

\displaystyle \\

Question 2: \displaystyle 5x^2+10x-3=0  

Answer:

Comparing \displaystyle 5x^2+10x-3=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 5, b = 10 \text{ and } c =-3  

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

\displaystyle \text{Therefore } x = \frac{-(10) \pm \sqrt{(10)^2-4(5)(-3)}}{2(5)}  

\displaystyle \text{Solving we get } x = 0.26, -2.26  

\displaystyle \\

Question 3: \displaystyle 2x^2-10x+5=0  

Answer:

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

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

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

\displaystyle \text{Solving we get } x = 4.44, 0.56  

\displaystyle \\

\displaystyle \text{Question 4: } 4x+ \frac{6}{x} +13=0  

Answer:

\displaystyle 4x+ \frac{6}{x} +13=0 or

\displaystyle 4x^2+13x+6=0  

Comparing \displaystyle 2x^2-10x+5=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 4, b = 13 \text{ and } c =6  

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

\displaystyle \text{Therefore } x = \frac{-(13) \pm \sqrt{(13)^2-4(4)(6)}}{2(4)}  

\displaystyle \text{Solving we get } x = -0.56, -2.69  

\displaystyle \\

Question 5: \displaystyle x^2-3x-9=0 [2007]

Answer:

Comparing \displaystyle x^2-3x-9=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 1, b = -3 \text{ and } c =-9  

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

\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-9)}}{2(1)}  

\displaystyle \text{Solving we get } x = 4.85, -1.85  

\displaystyle \\

Question 6: \displaystyle x^2-5x-10=0 [2013]

Answer:

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)}  

\displaystyle \text{Solving we get } x = 6.53, -1.53  

\displaystyle \\

Question 7: \displaystyle 3x^2-12x-1=0  

Answer:

Comparing \displaystyle 3x^2-12x-1=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 3, b = -12 \text{ and } c =-1  

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

\displaystyle \text{Therefore } x = \frac{-(-12) \pm \sqrt{(-12)^2-4(3)(-1)}}{2(3)}  

\displaystyle \text{Solving we get } x = 4.082, -0.082  

\displaystyle \\

Question 8: \displaystyle x^2-16x+6=0  

Answer:

Comparing \displaystyle x^2-16x+6=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 1, b = -16 \text{ and } c =6  

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

\displaystyle \text{Therefore } x = \frac{-(-16) \pm \sqrt{(-16)^2-4(1)(6)}}{2(1)}  

\displaystyle \text{Solving we get } x = 15.616, 0.384  

\displaystyle \\

Question 9: \displaystyle 2x^2+11x+4=0  

Answer:

Comparing \displaystyle 2x^2+11x+4=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 2, b = 11 \text{ and } c =4  

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

\displaystyle \text{Therefore } x = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(2)}  

\displaystyle \text{Solving we get } x = -0.392, -5.108  

\displaystyle \\

Question 10: \displaystyle x^4-2x^2-3=0  

Answer:

\displaystyle x^4-2x^2-3=0  

\displaystyle \text{Let } x^2 = y  

\displaystyle \text{Therefore } y^2-2y-3=0  

Comparing \displaystyle y^2-2y-3=0 with \displaystyle ay^2+by+c=0 , we get \displaystyle a = 1, b = -2 \text{ and } c =-3  

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

\displaystyle \text{Therefore } 7 = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}  

\displaystyle \text{Solving we get } y = 3, -1  

\displaystyle \text{Therefore } x^2 = 3\text{ or } x^2 = -1 (imaginary number).

\displaystyle \text{Hence } x = 1.732, -1.732\text{ or } \pm\sqrt{3}  

\displaystyle \\

Question 11: \displaystyle x^4-10x^2+9=0  

Answer:

\displaystyle x^4-10x^2+9=0  

\displaystyle \text{Let } y = x^2  

Therefore we have \displaystyle y^2-10y+9=0  

Comparing \displaystyle y^2-10y+9=0 with \displaystyle ay^2+by+c=0 , we get \displaystyle a = 1, b = -10 \text{ and } c =9  

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

\displaystyle \text{Therefore } 7 = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}  

\displaystyle \text{Solving we get } y = 9, 1  

\displaystyle \text{Therefore } x^2 = 9 \Rightarrow x = \pm 3\text{ or } x^2 = 1 \Rightarrow x = \pm 1 .

\displaystyle \text{Hence } x = \pm 3, \pm 1  

\displaystyle \\

Question 12: \displaystyle (x^2-x)^2+5(x^2-x)+4=0  

Answer:

\displaystyle \text{Let } (x^2-x)=y  

\displaystyle \text{Therefore } y^2+5y+4=0  

Comparing \displaystyle y^2+5y+4=0 with \displaystyle ay^2+by+c=0 , we get \displaystyle a = 1, b = 5 \text{ and } c =4  

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

\displaystyle \text{Therefore } y = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}  

\displaystyle \text{Solving we get } y = -1, -4  

\displaystyle \text{When } y = -1 \Rightarrow x^2-x+1=0  

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

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

\displaystyle \text{Hence } x = Imaginary  

\displaystyle \text{When } y = -4 \Rightarrow x^2-x+4=0  

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

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

\displaystyle \text{Hence } x = Imaginary  

\displaystyle \\

Question 13: \displaystyle (x^2-3x)^2-16(x^2-3x)-36=0  

Answer:

\displaystyle \text{Let } (x^2-3x)=y  

\displaystyle \text{Therefore } y^2-16y-36=0  

Comparing \displaystyle y^2-16y-36=0 with \displaystyle ay^2+by+c=0 , we get \displaystyle a = 1, b = -16 \text{ and } c =-36  

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

\displaystyle \text{Therefore } y = \frac{-(16) \pm \sqrt{(-16)^2-4(1)(-36)}}{2(1)}  

\displaystyle \text{Solving we get } y = 18, -2  

\displaystyle \text{When } y = 18 \Rightarrow x^2-3x-18=0  

Comparing \displaystyle x^2-3x-18=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 1, b = -3 \text{ and } c =-18  

\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-18)}}{2(1)}  

\displaystyle \text{Hence } x = 6, -3  

\displaystyle \text{When } y = -2 \Rightarrow x^2-x+2=0  

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

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

\displaystyle \text{Hence } x = 2, 1  

\displaystyle \text{Hence } x = 1, 2, 6, -3  

\displaystyle \\

\displaystyle \text{Question 14: } \sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}} = \frac{5}{2}  

Answer:

\displaystyle \sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}} = \frac{5}{2}  

\displaystyle \text{Let } \sqrt{\frac{x}{x-3}} = y  

Therefore the equation becomes

\displaystyle y + \frac{1}{y} = \frac{5}{2}  

or \displaystyle 2y^2-5y+2=0  

Comparing \displaystyle 2y^2-5y+2=0 with \displaystyle ay^2+by+c=0 , we get \displaystyle a = 2, b = -5 \text{ and } c =2  

\displaystyle \text{Therefore } y = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(c)}}{2(2)}  

\displaystyle \text{Hence } y = 2, 0.5  

When y = 2

\displaystyle \sqrt{\frac{x}{x-3}} = 2  

Squaring both sides

\displaystyle \frac{x}{x-3} = 4 \Rightarrow 3x=12 \Rightarrow x = 4  

Similarly, when y = 0.5

\displaystyle \sqrt{\frac{x}{x-3}} = 0.25 \Rightarrow 3x=-3 \Rightarrow x = -1  

\displaystyle \text{Hence } x = -1, 4  

\displaystyle \\

\displaystyle \text{Question 15: } ( \frac{2x-3}{x-1} )-4( \frac{x-1}{2x-3} )=3  

Answer:

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

\displaystyle \text{Let } ( \frac{2x-3}{x-1} ) = y  

\displaystyle \text{Therefore } y - \frac{4}{y} =3  

\displaystyle y^2-3y-4=0  

\displaystyle y^2-4y+y-4=0  

\displaystyle y(y-4)+(y-4)=0  

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

\displaystyle \text{When } y = 4  

\displaystyle \frac{2x-3}{x-1} =4  

\displaystyle 2x-3=4x-4 \Rightarrow x = 0.5  

\displaystyle \text{When } y = -1  

\displaystyle \frac{2x-3}{x-1} =-1  

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

\displaystyle \text{Hence } x = 0.5, \frac{4}{3}  

\displaystyle \\

\displaystyle \text{Question 16: } ( \frac{3x+1}{x+1} )+( \frac{x+1}{3x+1} )= \frac{5}{2}  

Answer:

\displaystyle ( \frac{3x+1}{x+1} )+( \frac{x+1}{3x+1} )= \frac{5}{2}  

\displaystyle \text{Let } ( \frac{3x+1}{x+1} ) = y  

\displaystyle y + \frac{1}{y} = \frac{5}{2}  

\displaystyle 2(y^2+1)=5y  

\displaystyle 2y^2-5y+2=0  

\displaystyle 2y^2-4y-y+2=0  

\displaystyle 2y(y-2)-1(y-2)=0  

\displaystyle (y-2)(2y-1)=0  

\displaystyle y = 2, 0.5  

\displaystyle \text{When } y = 2  

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

\displaystyle 3x+1=2x+2 \Rightarrow x =1  

\displaystyle \text{When } y =0.5  

\displaystyle ( \frac{3x+1}{x+1} ) = y  

\displaystyle 6x+2=x+1 \Rightarrow x = - \frac{1}{5}  

\displaystyle \text{Hence } x = 1, - \frac{1}{5}  

\displaystyle \\

\displaystyle \text{Question 17: } 3\sqrt{\frac{x}{5}} +3\sqrt{\frac{5}{x}} =10  

Answer:

\displaystyle 3\sqrt{\frac{x}{5}} +3\sqrt{\frac{5}{x}} =10  

\displaystyle \text{Let } \sqrt{\frac{x}{5}} = y  

\displaystyle 3y+ \frac{3}{y} =10  

\displaystyle 3y^2-10y+3=0  

\displaystyle 3y^2-9y-y+3=0  

\displaystyle 3y(y-3)-1(y-3)=0  

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

\displaystyle \text{When } y = 3  

\displaystyle \sqrt{\frac{x}{5}} = 3  

\displaystyle \frac{x}{5} =9 \Rightarrow x = 45  

\displaystyle \text{When } y = \frac{1}{3}  

\displaystyle \sqrt{\frac{x}{5}} = \frac{1}{3}  

\displaystyle \frac{x}{5} = \frac{1}{9} \Rightarrow x = \frac{5}{9}  

\displaystyle \text{Hence } x = 45, \frac{5}{9}  

\displaystyle \\

\displaystyle \text{Question 18: } 2x- \frac{1}{x} =7 [2006]

Answer:

\displaystyle 2x- \frac{1}{x} =7  

\displaystyle 2x^2-7x-1=0  

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

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

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

\displaystyle \text{Solving we get } x = 3.64, -0.14  

\displaystyle \\

Question 19: \displaystyle 5x^2-3x-4=0 [2012]

Answer:

Comparing \displaystyle 5x^2-3x-4=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 5, b = -3 \text{ and } c =-4  

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

\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(5)(-4)}}{2(5)}  

\displaystyle \text{Solving we get } x = 1.243, -0.643  

If we want only three significant figures than \displaystyle x = 1.24, -0.643  

\displaystyle \\

Question 20: \displaystyle (x-1)^2-3x+4=0 [2014]

Answer:

\displaystyle (x-1)^2-3x+4=0  

\displaystyle x^2+1-2x-3x+4=0  

\displaystyle x^2-5x+5=0  

Comparing \displaystyle 5x^2-3x-4=0 with \displaystyle ax^2+bx+c=0 , we get \displaystyle a = 1, b = -5 \text{ and } c =5  

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

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

\displaystyle \text{Solving we get } x = 3.618, 1.382  

If we want only two significant figures than \displaystyle x = 3.6, 1.3