Solve each of the following use quadratic formula

Question 1: $x^2-6x=27$

Comparing $x^2-6x=27$ with $ax^2+bx+c=0$, we get $a = 1, b = -6 \ and \ c =-27$

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

Therefore $x = \frac{-(-6) \pm \sqrt{(-6)^2-4(1)(-27)}}{2(1)}$

Solving we get $x = 9, -3$

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Question 2: $x^2-10x+21=0$

Comparing $x^2-10x+21=0$ with $ax^2+bx+c=0$, we get $a = 1, b = -10 \ and \ c =21$

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

Therefore $x = \frac{-(-10) \pm \sqrt{(-10)^2-4(1)(21)}}{2(1)}$

Solving we get $x = 3, 7$

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Question 3: $x^2+6x-10=0$

Comparing $x^2+6x-10=0$ with $ax^2+bx+c=0$, we get $a = 1, b = 6 \ and \ c =-10$

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

Therefore $x = \frac{-(6) \pm \sqrt{(6)^2-4(1)(-10)}}{2(1)}$

Solving we get $x = -3 \pm \sqrt{19}$

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Question 4: $x^2+2x-6=0$

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

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

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

Solving we get $x = -1 \pm \sqrt{7}$

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Question 5: $3x^2+2x-1=0$

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

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

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

Solving we get $x = -1, \frac{1}{3}$

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Question 6: $2x^2+7x+5=0$

Comparing $2x^2+7x+5=0$ with $ax^2+bx+c=0$, we get $a = 2, b = 7 \ and \ c =5$

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

Therefore $x = \frac{-(7) \pm \sqrt{(7)^2-4(2)(5)}}{2(2)}$

Solving we get $x = -1, -\frac{5}{2}$

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Question 7: $\frac{2}{3}x=-\frac{1}{6}x^2-\frac{1}{3}$

$\frac{2}{3}x=-\frac{1}{6}x^2-\frac{1}{3}$

Multiplying by 6

$4x = -x^2-2$ or $x^2+4x+2=0$

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

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

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

Solving we get $x = -2 \pm \sqrt{2}$

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Question 8: $\frac{1}{15}x^2+\frac{5}{3}=\frac{2}{3}x$

$\frac{1}{15}x^2+\frac{5}{3}=\frac{2}{3}x$

Multiplying by 15 and simplifying we get

$x^2+25=10x$ or $x^2-10x+25=0$

Comparing $x^2-10x+25=0$ with $ax^2+bx+c=0$, we get $a = 1, b = -10 \ and \ c =25$

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

Therefore $x = \frac{-(-10) \pm \sqrt{(-10)^2-4(1)(25)}}{2(1)}$

Solving we get $x = 5, 5$

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Question 9: $x^2-6x=2\sqrt{2}x$

$x^2-6x=2\sqrt{2}x$

or $x^2-2\sqrt{2}x-6x=0$

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

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

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

Solving we get $x = 3\sqrt{2}, -\sqrt{2}$

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Question 10: $\frac{4}{x}-3=\frac{5}{2x+3}$

$\frac{2x+3}{x+3} = \frac{x+4}{x+2}$

$(2x+3)(4-3x)=5x$

$8x+12-6x^2-9x = 5x$

$6x^2+6x-12=0$ or $x^2+x-2=0$

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

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

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

Solving we get $x = -2, 1$

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Question 11: $\frac{2x+3}{x+3} = \frac{x+4}{x+2}$

$\frac{2x+3}{x+3} = \frac{x+4}{x+2}$

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

$2x^2+7x+6 = x^2+7x+12$

$x^2-6=0$

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

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

Therefore $x = \frac{-(0) \pm \sqrt{(0)^2-4(1)(-6)}}{2(1)}$

Solving we get $x = \pm \sqrt{6}$

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Question 12: $\sqrt{6} x^2-4x-2\sqrt{6}=0$

$\sqrt{6} x^2-4x-2\sqrt{6}=0$

Comparing $\sqrt{6} x^2-4x-2\sqrt{6}=0$ with $ax^2+bx+c=0$, we get $a = \sqrt{6}, b = -4 \ and \ c =-2\sqrt{6}$

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

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

Solving we get $x = \sqrt{6}, -\frac{\sqrt{6}}{3}$

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Question 13: $\frac{2x}{x-4}+\frac{2x-5}{x-3}=8\frac{1}{3}$

$\frac{2x}{x-4}+\frac{2x-5}{x-3}=8\frac{1}{3}$

$3[2x(x-3)+(2x-5)(x-4)]=25(x-4)(x-3)$

$3(2x^2-6x+2x^2-13x+20) = 25(x^2-7x+12)$

$12x^2-57x+60=25x^2-175x+300$

$13x^2-118x+240=0$

Comparing $13x^2-118x+240=0$ with $ax^2+bx+c=0$, we get $a = 13, b = -118 \ and \ c =240$

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

Therefore $x = \frac{-(-118) \pm \sqrt{(-118)^2-4(13)(240)}}{2(13)}$

Solving we get $x = 6, \frac{40}{13}$

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Question 14: $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}$

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

$3[(x-1)(x-4)+(x-2)(x-3)]=10(x-2)(x-4)$

$3(2x^2-10x+10)=10(x^2-6x+8)$

$6x^2-30x+30=10x^2-60x+80$

$4x^2-30x+50=0 \ or \ 2x^2-15x+25=0$

Comparing $2x^2-15x+25=0$ with $ax^2+bx+c=0$, we get $a = 2, b = -15 \ and \ c =25$

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

Therefore $x = \frac{-(-15) \pm \sqrt{(-15)^2-4(2)(25)}}{2(2)}$

Solving we get $x = 5, \frac{5}{2}$

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Without solving, comment on the roots of each of the equations

Notes: The nature of the roots depends on the value of the $b^2-4ac$. Also in the equation $ax^2+bx+c=0$, $a, b, and c$ are real and $a \leq 0$.  The following cases can happen:

i) $b^2-4ac = 0$: In this case the roots would be real and equal.

ii) $b^2-4ac > 0$: In this case the roots would be real and unequal.

iii) $b^2-4ac < 0$: In this case the roots would be imaginary numbers

Question 15: $7x^2-9x+2=0$

Comparing $7x^2-9x+2=0$ with $ax^2+bx+c=0$, we get $a = 7, b = -9 \ and \ c =2$

Therefore $b^2-4ac = (-9)^2-4(7)(2) = 81 - 56 = 25 > 0$

Hence since $b^2-4ac > 0$: In this case the roots would be real and unequal.

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Question 16: $6x^2-13x+4=0$

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

Therefore $b^2-4ac = (-13)^2-4(6)(4) = 169 - 96 = 73 > 0$

Hence since $b^2-4ac > 0$: In this case the roots would be real and unequal.

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Question 17: $25x^2-10x+1=0$

Comparing $25x^2-10x+1=0$ with $ax^2+bx+c=0$, we get $a = 25, b = -10 \ and \ c =1$

Therefore $b^2-4ac = (-10)^2-4(25)(1) = 100 - 100 = 0$

Hence since $b^2-4ac = 0$: In this case the roots would be real and equal

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Question 18: $x^2+2\sqrt{3}-9=0$

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

Therefore $b^2-4ac = (-9)^2-4(1)(2\sqrt{3}) = 81 - 8\sqrt{3} > 0$

Hence since $b^2-4ac > 0$: In this case the roots would be real and unequal

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Question 19: $x^2-ax-b^2=0$

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

Therefore $b^2-4ac = (-a)^2-4(1)(-b^2) = a^2+b^2 > 0$

Hence since $b^2-4ac > 0$: In this case the roots would be real and unequal

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Question 20: $2x^2+8x+9=0$

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

Therefore $b^2-4ac = (-a)^2-4(1)(-b^2) = a^2+b^2 > 0$

Hence since $b^2-4ac > 0$: In this case the roots would be real and unequal

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Find the value of $p$ if the quadratic equations have equal roots:

Question 21: $4x^2-(p-2)x+1=0$

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

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

Substituting

$(-(p-2)^2)-4(4)(1) = 0$

$p^2+4-4p-16=0$

$p^2-4p-12=0$

$p^2-6p+2p-12=0$

$p(p-6)+2(p-=0$

$(p+2)(p-6)=0 \Rightarrow p = 6 or p = -2$

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Question 22: $x^2+(p-3)x+p=0$

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

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

Substituting

$((p-3)^2)-4(1)(p) = 0$

$p^2+9-6p-4p=0$

$p^2-10p+9=0$

$p^2-9p-p+9=0$

$(p-1)(p-9)=0 \Rightarrow p = 1 or p = 9$

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Question 23: $3x^2-12x+(p-5)=0$

$3x^2-12x+(p-5)=0$

Comparing $3x^2-12x+(p-5)=0$ with $ax^2+bx+c=0$, we get $a = 3, b = -12 \ and \ c =(n-5)$

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

$(-12)^2-4(3)(n-5)=0$

$144-12n+60=0$

$12n = 84$

or $n = 6$

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Question 24: $(p-2)x^2-(5+p)x+16=0$

$(p-2)x^2-(5+p)x+16=0$

Comparing $(p-2)x^2-(5+p)x+16=0$ with $ax^2+bx+c=0$, we get $a = (p-2), b = -(5+p) \ and \ c =16$

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

$(-(5+p)^2)-4(p-2)(16)=0$

$25+p^2+10p-64p+128=0$

$p^2-54p+153=0$

$p^2-51p-3p+153=0$

$p(p-51)-3(p-51)=0$

$(p-3)(p-51)=0 \Rightarrow p = 3 \ or \ 51$

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