Solve each of the following equations for $x$

Question 1:  $x^2-8x+5=0$

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

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: $2x^2-10x+5=0$

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

$4x+$ $\frac{6}{x}$ $+13=0$ 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]

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]

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: $3x^2-12x-1=0$

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: $x^2-16x+6=0$

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: $2x^2+11x+4=0$

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

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

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: $x^4-10x^2+9=0$

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

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

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

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

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

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

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

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

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

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

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

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]

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

$2x^2-7x-1=0$

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]

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]

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

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

$x^2-5x+5=0$

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 $x = 3.6, 1.3$