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Algebra AMC 10 Math Olympiad USA Math Olympiad

Sum of reciprocals Problem | AMC-10A, 2003 | Problem 18

Try this beautiful problem from Algebra based on Sum of reciprocals in the quadratic equation.

Sum of reciprocals in equation – AMC-10A, 2003- Problem 18


What is the sum of the reciprocals of the roots of the equation \(\frac{2003}{2004} x^2 +1+\frac{1}{x}=0\)?

  • \(\frac{1}{2}\)
  • \(-1\)
  • \(-\frac{1}{2}\)

Key Concepts


Algebra

quadrratic equation

root of the equation

Check the Answer


Answer: \(-1\)

AMC-10A (2003) Problem 18

Pre College Mathematics

Try with Hints


The given equation is \(\frac{2003}{2004} x^2 +1+\frac{1}{x}=0\).after simplification we will get,

\(2003 x^2 +2004 x+2004=0\) which is a quadratic equation .we have to find out the roots of the equation.

suppose there is a quadratic equation \(ax^2 +bx+c=0\) (where a,b,c are constant) and roots of the equation are \(p_1\) & \(p_2\) then we know that

\(p_1 +p_2\)=-\(\frac{b}{a}\) & \(p_1 p_2\)=\(\frac{c}{a}\).now can you find out sum of the reciprocals of the roots of the given equation..?

can you finish the problem……..

The given equation is \(2003 x^2 +2004 x+2004=0\).Let \(m_1\) &\(m_2\) are the roots of the given equation

Then \(m_1 +m_2\)=-\(\frac{2004}{2003}\) & \(m_1 m_2=\frac{2004}{2003}\).Now can you find out sum of the reciprocals of the roots of the given equation?

can you finish the problem……..

Now \(\frac{m_1 +m_2}{m_1m_2}\)=\(\frac{1}{m_1} +\frac{1}{m_2}\)=\(\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}\)=\(-1\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Quadratic equation Problem | AMC-10A, 2003 | Problem 5

Try this beautiful problem from Algebra based on quadratic equation.

Quadratic equation – AMC-10A, 2003- Problem 5


Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$?

  • \(2\)
  • \(0\)
  • \(\frac{7}{2}\)

Key Concepts


algebra

quadratic equation

Roots

Check the Answer


Answer: \(0\)

AMC-10A (2003) Problem 5

Pre College Mathematics

Try with Hints


To find out the value of \((d-1)(e-1)\),at first we have to find out the value of \(d\) and \(e\).Given that \(d\) and \(e\) are the solutions of the equations $2x^{2}+3x-5=0$ that means \(d\) and \(e\) are the roots of the given equation.so if we find out the values of roots from the given equation then we will get \(d\) and \(e\).Can you find out the roots?

Can you now finish the problem ……….

To find out the roots :

The given equation is \(2x^{2}+3x-5=0\) \(\Rightarrow (2x+5)(x-1)=0\) \(\Rightarrow x=1 or \frac{-5}{2}\)

Therefore the values of \(d\) and \(e\) are \(1\) and \(\frac{-5}{2}\) respectively

can you finish the problem……..

Therefore \((d-1)(e-1)\)=\((1-1)(\frac{-5}{2} -1)\)=\(0\)

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Algebra AMC 10 Math Olympiad USA Math Olympiad

Problem on Equation | AMC-10A, 2007 | Problem 20

Try this beautiful problem from Algebra based on quadratic equation

Problem on Equation – AMC-10A, 2007- Problem 20


Suppose that the number \(a\) satisfies the equation \(4 = a + a^{ – 1}\). What is the value of \(a^{4} + a^{ – 4}\)?

  • \(174\)
  • \(194\)
  • \(156\)

Key Concepts


Algebra

Linear equation

multiplication

Check the Answer


Answer: \(194\)

AMC-10A (2007) Problem 20

Pre College Mathematics

Try with Hints


Given that \(4 = a + a^{ – 1}\). we have to find out the value \(a^{4} + a^{ – 4}\)

Squarring both sides of \(a^{4} + a^{ – 4}\) …then opbtain…

can you finish the problem……..

\((a + a^{ – 1})^2=4^2\) \(\Rightarrow (a^2 + a^{-2} +2)=16\) \(\Rightarrow a^2 + a^{-2}=14\) and now squarring both side again………….

can you finish the problem……..

Squarring both sides of \(a^2 + a^{-2}=14\) \(\Rightarrow (a^2 + a^{-2})^2=(14)^2\) \(\Rightarrow a^4 + a^{-4} +2=196\) \(\Rightarrow a^4 + a^{-4}=194\)

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India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Sum of two digit numbers | PRMO-2016 | Problem 7

Try this beautiful problem from Algebra based on Sum of two digit numbers from PRMO 2016.

Sum of two digit numbers | PRMO | Problem 7


Let s(n) and p(n) denote the sum of all digits of n and the products of all the digits of n(when written in decimal form),respectively.Find the sum of all two digits natural numbers n such that \(n=s(n)+p(n)\)

  • $560$
  • $531$
  • $654$

Key Concepts


Algebra

number system

addition

Check the Answer


Answer:$531$

PRMO-2016, Problem 7

Pre College Mathematics

Try with Hints


Let \(n\) is a number of two digits ,ten’s place \(x\) and unit place is \(y\).so \(n=10x +y\).given that \(s(n)\)= sum of all digits \(\Rightarrow s(n)=x+y\) and \(p(n)\)=product of all digits=\(xy\)

now the given condition is \(n=s(n)+p(n)\)

Can you now finish the problem ……….

From \(n=s(n)+p(n)\) condition we have,

\(n=s(n)+p(n)\) \(\Rightarrow 10x+y=x+y+xy \Rightarrow 9x=xy \Rightarrow y=9\) and the value of\(x\) be any digit….

Can you finish the problem……..

Therefore all two digits numbers are \(19,29,39,49,59,69,79,89,99\) and sum=\(19+29+39+49+59+69+79+89+99=531\)

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Algebra AMC 10 Math Olympiad Theory of Equations USA Math Olympiad

Quadratic Equation Problem | AMC-10A, 2005 | Problem 10

Try this beautiful problem from Algebra based on Quadratic Equation….

Quadratic equation – AMC-10A, 2005- Problem 10


There are two values of $a$ for which the equation $4 x^{2}+a x+8 x+9=0$ has only one solution for $x$. What is the sum of those values of $a$ ?

  • \(5\)
  • \(20\)
  • \(-16\)
  • \(25\)
  • \(36\)

Key Concepts


algebra

Quadratic equation

Equal roots

Check the Answer


Answer: \(-16\)

AMC-10A (2005) Problem 10

Pre College Mathematics

Try with Hints


The given equation is $4 x^{2}+a x+8 x+9=0$

\(\Rightarrow 4 x^{2}+x(a+8)+9=0\)

comparing the above equation with \(Ax^2-Bx+C=0\) we will get \(A=4\),\(B=(a+8)\),\(C=9\)

Now for equal roots of a quadratic equation \(B^2-4Ac=0\)

Can you now finish the problem ……….

Now \(B^2-4Ac=0\) becomes

\((a+8)^2-4\times 9 \times 4=0\)

\(\Rightarrow (a+8)^2=144\)

\(\Rightarrow (a+8)=\pm 12\)

\(\Rightarrow a=+4 \) & \(-20\)

Therefore The sum of the values of \(a=-20+4=-16\)

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I.S.I. and C.M.I. Entrance

Quadratic equation | ISI-B.stat | Objective Problem 198

Try this beautiful problem Based on Quadratic equation, useful for ISI B.Stat Entrance.

Quadratic equation | ISI B.Stat Entrance | Problem 198


Consider the quadratic equation of the form \(x^2 + bx + c = 0\). The number of such equations that have real roots and coefficients b and c from the set \(\{1, 2, 3, 4, 5\}\) (b and c may be equal) is

  • (a) \(18\)
  • (b) \(15)\)
  • (c) \(12\)
  • (d) None of these

Key Concepts


Quadratic equation

Algebra

Roots of the nature

Check the Answer


Answer: (c) \(12 \)

TOMATO, Problem 198

Challenges and Thrills in Pre College Mathematics

Try with Hints


The given equation is \(x^2 + bx + c = 0\).

we know that in the equation \(ax^2+bx+c=0\) ,the condition for real root is \(b^2-4ac \geq 0\)

Can you now finish the problem ……….

Now, \(b^2 > 4c\)
b cannot be equal to 1.
If b = 2, c = 1
If b = 3, c = 1, 2
If b = 4, c = 1, 2, 3, 4
If b = 5, c = 1, 2, 3, 4, 5
Total number of equations = \(1 + 2 + 4 + 5 = 12\)

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AMC-8 India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Roots of Equations | PRMO-2016 | Problem 8

Try this beautiful problem from Algebra based on roots of equations.

Roots of Equations | PRMO | Problem 8


Suppose that \(a\) and \(b\) are real numbers such that \(ab \neq 1\) and the equations \(120 a^2 -120a+1=0\) and \(b^2-120b+120=0\) hold. Find the value of \(\frac{1+b+ab}{a}\)

  • $200$
  • $240$
  • $300$

Key Concepts


Algebra

quadratic equation

Roots

Check the Answer


Answer:$240$

PRMO-2016, Problem 8

Pre College Mathematics

Try with Hints


The given equations are \(120 a^2 -120a+1=0\) and \(b^2-120b+120=0\).we have to find out the values of \(a\) and \(b\)….

Let \(x,y\) be the roots of the equation \(120 a^2 -120a+1=0\)then \(\frac{1}{x},\frac{1}{y}\) be the roots of the equations of \(b^2-120b+120=0\).can you find out the value of \(a\) & \(b\)

Can you now finish the problem ……….

From two equations after sim[lificatiopn we get…\(a=x\) and \(b=\frac{1}{y}\) (as \(ab \neq 1)\)

Can you finish the problem……..

\(\frac{1+b+ab}{a}\)=\(\frac { 1+\frac{1}{y} +\frac{x}{y}}{x}\)=\(\frac{(x+y)+1}{xy}=240\)

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Algebra Math Olympiad PRMO

Problem based on Integer | PRMO-2018 | Problem 6

Try this beautiful problem from Algebra based on Integer

Algebra based on Integer PRMO Problem 6


Integers a, b, c satisfy \(a + b – c = 1\) and \(a^2 + b^2 – c^2 = –1\). What is the sum of all possible values of
\(a^2 + b^2 + c^2\) ?

  • $24$
  • $18$
  • $34$

Key Concepts


Algebra

quadratic equation

Factorization

Check the Answer


Answer:$18$

PRMO-2018, Problem 6

Pre College Mathematics

Try with Hints


Use \((a + b – 1 )= c\) this relation

Can you now finish the problem ……….

\((a + b – 1 )^2= c^2\) (squaring both sides…….)

Can you finish the problem……..

Given that a,b,c are integer satisfy \(a + b – c = 1\) and \(a^2 + b^2 – c^2 = –1\).

Now

\((a + b – c )= 1\)

\(\Rightarrow (a + b – 1 )= c\)

\(\Rightarrow (a + b – 1 )^2= c^2\) (squarring both sides…….)

\(\Rightarrow (a^2 + b^2 +1^2+2ab-2a-2b)= c^2\)

\(\Rightarrow (a^2 + b^2 –c^2+1+2ab)= 2(a+b)\)

\(\Rightarrow (-1)+1+2ab= 2(a+b)\) ( as \(a^2 + b^2 – c^2 = –1\).

\(\Rightarrow ab=a+b\)

\(\Rightarrow (a-1)(b-1)=1\)

Therefore the possibile cases are \((a-1)=\pm 1\) and \((b-1)=\pm 1\)

Therefore \( a=1\),\(b=1\) or \(a=0\),\(b=0\)

From the equation \((a + b – c )= 1\) ,C =3 (as a=b=2) and C=-1(as a=b=0)

Therefore \((a^2 +b^2 +c^2)=4+4+9=17\) and \((a^2 +b^2 +c^2)=0+0+1=1\)

sum of all possible values of \(a^2 + b^2 + c^2=17+1=18\)

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Algebra Math Olympiad PRMO

Quadratic Equation Problem | PRMO-2018 | Problem 9

Try this beautiful problem from Algebra based on Quadratic equation.

Algebra based on Quadratic equation PRMO Problem 9


Suppose a,b are integers and a + b is a root of \(x^2 +ax+b=0\).What is the maximum possible
values of \( b^2 \)?

  • $49$
  • $81$
  • $64$

Key Concepts


Algebra

quadratic equation

Factorization

Check the Answer


Answer:$81$

PRMO-2018, Problem 9

Pre College Mathematics

Try with Hints


(‘a+b”) is the root of the equation therefore (“a+b”) must satisfy the given equation

Can you now finish the problem ……….

Discriminant is a perfect square

can you finish the problem……..

Given that \(‘a+b”\) is the root of the equation therefore \(“a+b”\) must satisfy the given equation

Therefore the given equation becomes ……

\((a+b)^2 +a(a+b)+b=0\)

\(\Rightarrow a^2 +2ab+b^2+a^2+ab+b=0\)

\(\Rightarrow 2a^2 +3ab+b^2+b=0\)

Now since “a” is an integer,Discriminant is a perfect square

\(\Rightarrow 9b^2 -8(b^2+b)=m^2\) (for some \(m \in \mathbb Z)\)

\(\Rightarrow (b-4)^2 -16=m^2\)

\(\Rightarrow (b-4+m)(b-4-m)=16\)

Therefore the possible cases are  \(b-4+m=\pm 8\), \(b-4-m=\pm 2\),\(b-4+m=b-4-m=\pm 4\)

 i.e b-4=5,-5,4,-4

\(\Rightarrow b =9,-1,8,0\)

 Therefore \( (b^2)_{max} = 81\)

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Algebra AMC-8 Math Olympiad USA Math Olympiad

Quadratic equation Problem | AMC 8, 2009 | Problem 23

Try this beautiful problem from Algebra based on Quadratic equation.

Algebra based on Quadratic equation – AMC-8, 2009 – Problem 23


On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought  400 jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

  • $34$
  • $28$
  • $25$

Key Concepts


Algebra

quadratic equation

Factorization

Check the Answer


Answer:$28$

AMC-8 (2009) Problem 23

Pre College Mathematics

Try with Hints


Let the number of girls be x

so the number of boys be x+2

Can you now finish the problem ……….

She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

can you finish the problem……..

Let the number of girls be x

so the number of boys be x+2

she gave each girl x jellybeans and each boy x+2 jellybeans,

Therefore she gave total number of  jelly beans to girls be \(x^2\)

Therefore she gave total number of  jelly beans to boys be \((x+2)^2\)

 She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

\(\Rightarrow x^2 + x^2 +4x+4=394\)

\(\Rightarrow 2x^2 +4x-390=0\)

\(\Rightarrow x^2 +2x -195=0\)

\((x+15)(x-13)=0\)

i.e x=-15 , 13 (we neglect negetive as number of students can not be negetive )

Therefore x=13 i.e number os girls be 13 and number of boys be 13+2=15

total number of students ne 13+15=28

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