Solve each of the following use quadratic formula

Question 1:  x^2-6x=27  

Answer:

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

\\

Question 2: x^2-10x+21=0  

Answer:

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

\\

Question 3: x^2+6x-10=0  

Answer:

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}

\\

Question 4: x^2+2x-6=0  

Answer:

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}

\\

Question 5: 3x^2+2x-1=0  

Answer:

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}

\\

Question 6: 2x^2+7x+5=0  

Answer:

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}

\\

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

Answer:

\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}

\\

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

Answer:

\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

\\

Question 9: x^2-6x=2\sqrt{2}x  

Answer:

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}

\\

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

Answer:

\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

\\

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

Answer:

\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}

 

\\

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

Answer:

\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}

\\

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

Answer:

\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}

\\

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

Answer:

\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}

\\

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  

Answer:

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.

\\

Question 16: 6x^2-13x+4=0  

Answer:

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.

\\

Question 17: 25x^2-10x+1=0  

Answer:

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

\\

Question 18: x^2+2\sqrt{3}-9=0  

Answer:

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

\\

Question 19: x^2-ax-b^2=0  

Answer:

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

\\

Question 20: 2x^2+8x+9=0  

Answer:

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

\\

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

Question 21: 4x^2-(p-2)x+1=0  

Answer:

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  

\\

Question 22: x^2+(p-3)x+p=0  

Answer:

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  

 

\\

Question 23: 3x^2-12x+(p-5)=0  

Answer:

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

\\

Question 24: (p-2)x^2-(5+p)x+16=0  

Answer:

(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

\\

Advertisements