from: category_eng

Manipulation

Real numbers $x$ and $y$ satisfy the equation $x^2 + y^2 = 10x - 6y - 34$ . What is $x + y$ ?

$extbf{(A)} 1 qquad extbf{(B)} 2 qquad extbf{(C)} 3 qquad extbf{(D)} 6 qquad extbf{(E)} 8$

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 $x$ ?

$extbf{(A) } 1qquad extbf{(B) } 2qquad extbf{(C) } 3qquad extbf{(D) } 4qquad extbf{(E) } 5$

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

$mathrm{(A) } -frac{5}{2}qquad mathrm{(B) } 0qquad mathrm{(C) } 3qquad mathrm{(D) } 5qquad mathrm{(E) } 6$

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$ . What is $a+b$ ?

$ext{(A) }1qquad ext{(B) }2qquad ext{(C) }sqrt 5qquad ext{(D) }sqrt 6qquad ext{(E) }3$

10.

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 $n/p$ ?

$mathrm{(A)} {{{1}}} qquad mathrm{(B)} {{{2}}} qquad mathrm{(C)} {{{4}}} qquad mathrm{(D)} {{{8}}} qquad mathrm{(...$

11.

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

$mathrm{(A) } -16qquad mathrm{(B) } -8qquad mathrm{(C) } 0qquad mathrm{(D) } 8qquad mathrm{(E) } 20$

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5. 0	Understanding

6.

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 $y = -3$ . $x+y = 5 + (-3) = oxed{ extbf{(B) }2}$

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 $k=oxed{5}$ Â we 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.

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 $left(-frac{5}{2}-1<br /> ight)(1-1)=left(-frac{7}{2}<br /> ight)(0)=oxed{mathrm{(B)} 0}$

Solution 2

We can use the sum and product of a quadratic:

$(d-1)(e-1)=de-(d+e)+1 implies ext{product}- ext{sum}+1 implies dfrac{c}{a}-left(-dfrac{b}{a}<br /> ight)+1 implies dfrac{...$

Each of the numbers $a$ and $b$ is a solution to $left| x - frac 1x <br /> ight| = 1$ .

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 $frac{-1+sqrt 5}2 + frac{1+sqrt 5}2 = oxed{(C) sqrt 5}$ .

10.

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 $x^2 + 8x + 16 = 0, x^2 + 16x + 64 = 0$ .

11.

A quadratic equation has exactly one root if and only if it is a perfect square. So set

$4x^2 + ax + 8x + 9 = (mx + n)^2$

$4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2$

Two polynomials are equal only if their coefficients are equal, so we must have

$m^2 = 4, n^2 = 9$

$m = pm 2, n = pm 3$

$a + 8= 2mn = pm 2cdot 2cdot 3 = pm 12$

$a = 4$ or $a = -20$ .

So the desired sum is $(4)+(-20)=-16 Longrightarrow mathrm{(A)}$

Alternatively, note that whatever the two values of $a$ are, they must lead to equations of the form $px^2 + qx + r =0$ and $px^2 - qx + r = 0$ . So the two choices of $a$ must make $a_1 + 8 = q$ and $a_2 + 8 = -q$ so $a_1 + a_2 + 16 = 0$ and $a_1 + a_2 = -16Longrightarrow mathrm{(A)}$ .

Alternate Solution

Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have
$(a+8)^2 - 4(4)(9) = 0 implies a^2 + 16a - 144.$ We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the $pm$ sign when added). So we must have
$frac{-16 + sqrt{ ext{something}}}{2} + frac{-16 - sqrt{ ext{something}}}{2}.$
Therefore, we have $(-16)(2)/2 = -16 implies oxed{A}.$