Solve each of the following equations for x

Question 1:  x^2-8x+5=0

Answer:

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

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

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

Solving we get x = 7.32, 0.68

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Question 2: 5x^2+10x-3=0  

Answer:

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

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

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

Solving we get x = 0.26, -2.26

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

Answer:

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

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

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

Solving we get x = 4.44, 0.56

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

Answer:

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

4x^2+13x+6=0

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

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

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

Solving we get x = -0.56, -2.69

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Question 5: x^2-3x-9=0     [2007]

Answer:

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

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

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

Solving we get x = 4.85, -1.85

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Question 6: x^2-5x-10=0     [2013]

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

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

Answer:

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

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

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

Solving we get x = 4.082, -0.082

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

Answer:

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

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

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

Solving we get x = 15.616, 0.384

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

Answer:

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

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

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

Solving we get x = -0.392, -5.108

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

Answer:

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

Let x^2 = y

Therefore y^2-2y-3=0

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

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

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

Solving we get y = 3, -1

Therefore x^2 = 3 \ or \  x^2 = -1  (imaginary number).

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

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Question 11: x^4-10x^2+9=0

Answer:

x^4-10x^2+9=0

Let y = x^2 

Therefore we have y^2-10y+9=0

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

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

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

Solving we get y = 9, 1

Therefore x^2 = 9 \Rightarrow x = \pm 3 \ or\  x^2 = 1 \Rightarrow x = \pm  1 .

Hence x = \pm 3, \pm  1

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

Answer:

Let (x^2-x)=y

Therefore y^2+5y+4=0

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

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

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

Solving we get y = -1, -4

When  y = -1 \Rightarrow x^2-x+1=0 

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

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

Hence x = Imaginary 

When  y = -4 \Rightarrow x^2-x+4=0 

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

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

Hence x = Imaginary 

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

Answer:

Let (x^2-3x)=y

Therefore y^2-16y-36=0

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

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

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

Solving we get y = 18, -2

When  y = 18 \Rightarrow x^2-3x-18=0 

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

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

Hence x = 6, -3

When  y = -2 \Rightarrow x^2-x+2=0 

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

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

Hence x = 2, 1

Hence x = 1, 2, 6, -3

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

Answer:

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

Let \sqrt{\frac{x}{x-3}} = y

Therefore the equation becomes

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

or 2y^2-5y+2=0

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

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

Hence y = 2, 0.5

When y = 2

\sqrt{\frac{x}{x-3}}  = 2  

Squaring both sides

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

Similarly, when y = 0.5

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

Hence x = -1, 4

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

Answer:

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

Let (\frac{2x-3}{x-1}) = y

Therefore  y -\frac{4}{y}=3

y^2-3y-4=0

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

y(y-4)+(y-4)=0

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

When y = 4

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

2x-3=4x-4  \Rightarrow x = 0.5

When y = -1

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

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

Hence x = 0.5, \frac{4}{3}

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

Answer:

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

Let (\frac{3x+1}{x+1}) = y

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

2(y^2+1)=5y

2y^2-5y+2=0

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

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

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

y = 2, 0.5

When y = 2

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

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

When y =0.5

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

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

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

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

Answer:

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

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

3y+\frac{3}{y} =10

3y^2-10y+3=0

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

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

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

When y = 3

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

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

When y = \frac{1}{3}

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

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

Hence x = 45, \frac{5}{9}

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

Answer:

2x-\frac{1}{x}=7

2x^2-7x-1=0

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

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

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

Solving we get x = 3.64, -0.14

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Question 19: 5x^2-3x-4=0      [2012]

Answer:

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

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

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

Solving we get x = 1.243, -0.643

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

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

Answer:

(x-1)^2-3x+4=0

x^2+1-2x-3x+4=0

x^2-5x+5=0

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

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

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

Solving we get x = 3.618, 1.382

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

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