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Equation - Quadratic Equation

For COMPETITION
Number of Total Problems: 11.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: None

Section:Equation

Theme:None
Adjustment# :

Difficulty: 1

Category Quadratic Equation

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: None

Section:Equation

Theme:None
Adjustment# :

Difficulty: 1

Category Quadratic Equation

Analysis

Solution/Answer

Problem Num : 3

From : NCTM

Type: None

Section:Equation

Theme:None
Adjustment# :

Difficulty: 1

Category Quadratic Equation

Analysis

Solution/Answer

Problem Num : 4

From : NCTM

Type: None

Section:Equation

Theme:None
Adjustment# :

Difficulty: 2

Category Quadratic Equation

Analysis

Solution/Answer

Problem Num : 5

From : NCTM

Type: Understanding

Section:Equation

Theme:Manipulation
Adjustment# : 0

Difficulty: 1

Category Quadratic Equation

Analysis

Solution/Answer

Problem Num : 6

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
Real numbers $x$ and $y$ satisfy the equation $x^2 + y^2 = 10x - 6y - 34$ . What is $x + y$ ?
$extbf{(A)} 1 qquad extbf{(B)} 2 qquad extbf{(C)} 3 qquad extbf{(D)} 6 qquad extbf{(E)} 8$
'

Category Quadratic Equation

Analysis

Solution/Answer
If we complete the square after bringing the $x$ and $y$ terms to the other side, we get $(x-5)^2 + (y+3)^2 = 0$ . Squares of real numbers are nonnegative, so we need both $(x-5)^2$ and $(y+3)^2$ to be $0$ . This obviously only happens when $x = 5$ and $y = -3$ . $x+y = 5 + (-3) = oxed{ extbf{(B) }2}$

Answer:

Problem Num : 7

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
For which of the following values of $k$ does the equation $frac{x-1}{x-2} = frac{x-k}{x-6}$ have no solution for $x$ ?
$extbf{(A) } 1qquad extbf{(B) } 2qquad extbf{(C) } 3qquad extbf{(D) } 4qquad extbf{(E) } 5$

'

Category Quadratic Equation

Analysis

Solution/Answer
The domain over which we solve the equation is $mathbb{R} - {2,6}$ .
We can now cross-multiply to get rid of the fractions, we get $(x-1)(x-6)=(x-k)(x-2)$ .
Simplifying that, we get $7x-6 = (k+2)x - 2k$ . Clearly for $k=oxed{5}$ 횂혻we get the equation $-6=-10$ which is never true.
For other $k$ , one can solve for $x$ : $x(5-k) = 6-2k$ , hence $x=frac {6-2k}{5-k}$ . We can easily verify that for none of the other four possible values of $k$ is this equal to $2$ or $6$ , hence there is a solution for $x$ in each of the other cases.

Answer:

Problem Num : 8

From : AMC10

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty:

'Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$ ?
$mathrm{(A) } -frac{5}{2}qquad mathrm{(B) } 0qquad mathrm{(C) } 3qquad mathrm{(D) } 5qquad mathrm{(E) } 6$

Category Quadratic Equation

Analysis

Solution/Answer
Solution 1

Using factoring:
$2x^{2}+3x-5=0$
$(2x+5)(x-1)=0$
$x = -frac{5}{2}$ or $x=1$
So $d$ and $e$ are $-frac{5}{2}$ and $1$ .
Therefore the answer is $left(-frac{5}{2}-1 ight)(1-1)=left(-frac{7}{2} ight)(0)=oxed{mathrm{(B)} 0}$

Solution 2

We can use the sum and product of a quadratic:
$(d-1)(e-1)=de-(d+e)+1 implies ext{product}- ext{sum}+1 implies dfrac{c}{a}-left(-dfrac{b}{a} ight)+1 implies dfrac{...$

Answer:

Problem Num : 9

From : AMC12

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$ . What is $a+b$ ?
$ext{(A) }1qquad ext{(B) }2qquad ext{(C) }sqrt 5qquad ext{(D) }sqrt 6qquad ext{(E) }3$
'

Category Quadratic Equation

Analysis

Solution/Answer
Each of the numbers $a$ and $b$ is a solution to $left| x - frac 1x ight| = 1$ .
Hence it is either a solution to $x - frac 1x = 1$ , or to $frac 1x - x = 1$ . Then it must be a solution either to $x^2 - x - 1 = 0$ , or to $x^2 + x - 1 = 0$ .
There are in total four such values of $x$ , namely $frac{pm 1 pm sqrt 5}2$ .
Out of these, two are positive: $frac{-1+sqrt 5}2$ and $frac{1+sqrt 5}2$ . We can easily check that both of them indeed have the required property, and their sum is $frac{-1+sqrt 5}2 + frac{1+sqrt 5}2 = oxed{(C) sqrt 5}$ .

Answer:

Problem Num : 10

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$ ?
$mathrm{(A)} {{{1}}} qquad mathrm{(B)} {{{2}}} qquad mathrm{(C)} {{{4}}} qquad mathrm{(D)} {{{8}}} qquad mathrm{(...$
'

Category Quadratic Equation

Analysis

Solution/Answer
Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then
$x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,$
so $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so
$x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,$
and $m = -2(a+b)$ and $n = 4ab$ . Thus $frac{n}{p} = frac{4ab}{-(a+b)} = frac{4m}{frac{m}{2}} = oxed{mathrm{(D)} 8}$ .
Indeed, consider the quadratics $x^2 + 8x + 16 = 0, x^2 + 16x + 64 = 0$ .

Answer:

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