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Let a,b,c, and d be nonzero real numbers. If the quadratic equation

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Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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26 Apr 2016, 16:39
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$

B. $$-\frac{ac}{bd}$$

C. $$-\frac{ad}{bc}$$

D. $$\frac{ab}{cd}$$

E. $$\frac{ad}{bc}$$
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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26 Apr 2016, 19:37
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amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

Hi,
one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation..
$$ax(cx+d)=-b(cx+d)$$ ..
$$acx^2+(ad+bc)x+bd = 0$$..
Let the roots be S and T
Sum of roots = $$S+T = -(\frac{ad+bc}{ac})$$..
Product of roots = $$S*T = \frac{bd}{ac}$$..

two methods..

1) equate from SUM..
we know $$S+T = -(\frac{ad+bc}{ac})$$..
$$S+T = -(\frac{ad}{ac}+\frac{bc}{ac})$$
let $$S = -\frac{ad}{ac}$$and $$T = -\frac{bc}{ac}$$
so$$\frac{S}{T} = \frac{ad}{bc}$$
E

2)we are to find$$\frac{S}{T} or \frac{T}{S}$$..
divide SUM by PRODUCT--

$$\frac{S+T}{ST} = -(\frac{ad+bc}{ac})/\frac{bd}{ac}$$..
$$\frac{S}{ST}+\frac{T}{ST} = -( \frac{ad}{bd} + \frac{bc}{bd})$$ ..
$$\frac{1}{T} + \frac{1}{S} = -\frac{a}{b} - \frac{c}{d}$$..

so JUST for finding the ratio, we will equate two sides--
Here it doesn't matter what do you equate with, because we are just finding any of the ratio

$$\frac{1}{T}= -\frac{a}{b}$$and $$\frac{1}{S}= -\frac{c}{d}$$..

$$\frac{\frac{1}{T}}{\frac{1}{S}}$$ = $$\frac{-\frac{a}{b}}{-\frac{c}{d}}$$..

$$\frac{S}{T} = \frac{ad}{bc}$$

E
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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26 Apr 2016, 20:19
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amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

$$ax(cx+d)=-b(cx+d)$$

You solve a quadratic by splitting it into factors.

$$ax(cx+d) + b(cx+d) = 0$$

But recognise that the factors are already there in this equation. Take (cx + d) common.
$$(ax + b)*(cx + d) = 0$$

Roots are -b/a and -d/c

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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02 May 2016, 18:40
Don't Even Think about calculating the roots by opening the brackets
It becomes a mess..!
Here take cx+d common and solve for x
Hence E
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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05 May 2016, 01:58
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dgeorgie wrote:
Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx+d)=-b(cx+d) is solved for x, which of the following is a possible ration of the 2 solutions?

A. -ab/(cd)
B. -ac/(bd)
D. ab/(cd)

Actually, I dont understand why do we have a quadratic equation. Isn't it the same as ax=-b ? Then we would have only 1 solution

No, you cannot cancel (cx+d) from both the sides. Because (cx+d) can be 0 too and you cannot cancel 0 from both sides.
You can easily cancel out any numbers, but not variables.

For Example: x(x+2) = 5(x+2). In this case, you cannot simply cancel out (x+2) from both the sides.

Therefore ax(cx+d)=-b(cx+d) should be written as (ax+b)(cx+d) = 0
Now in the quadratic equation, any of the terms can be equal to zero.

Therefore, the two solutions will be x = -b/a and x = -d/c

Ratio of two roots = (-d/c) / (-b/a) = ad/bc
Correct option: E

Does this help?

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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25 Jul 2016, 23:47
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Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx+d)=-b(cx+d) is solved for x, which of the following is a possible ration of the 2 solutions?

$$Let\mbox{'}s\; say\; a=1,\; b=2,\; c=3\; and\; d=4$$

the solutions would be:
a) -$$\frac{1}{6}$$
b) -$$\frac{3}{8}$$
c) -$$\frac{2}{3}$$
d) $$\frac{1}{6}$$
e) $$\frac{2}{3}$$

$$Now\; let\mbox{'}s\; solve\; the\; equation,\; after\; substituting\; a=1,\; b=2,\; c=3\; and\; d=4\; into\; the\; eq.\; we\; find: 3x^{2}\; +\; 10x\; +\; 8\; =\; 0$$
$$which\; solved\; gives\; us\; \left( x+\frac{4}{3} \right)\left( x+2 \right),\; where\; the\; roots\; are\; x_{a}=\; -\frac{4}{3}\; and\; x_{b}=\; -2,\; if\; we\; divide\; them:\; \frac{-\frac{4}{3}}{-2}\; we\; find\; \frac{xa}{xb}\; =\; \frac{2}{3}$$

$$Answer\; E$$

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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03 Aug 2016, 09:02
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amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

ax(cx + d) = -b(cx + d)

ax(cx + d) + b(cx + d) = 0

Next we can factor out (cx + d):

(cx + d)(ax + b) = 0

Using the zero product property, we can set the expression in each set of parentheses to zero:

cx + d = 0

cx = -d

x = -d/c

Or

ax + b = 0

ax = -b

x = -b/a

So a possible ratio is:

(-d/c)/(-b/a)

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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06 Aug 2016, 02:06
Hello, can someone please explain why this works:
ax(cx + d) = -b(cx + d)
axc^2 + axd = -bcx - bd
ax^2 + axd + bcx + bd = 0
ax^2 + (ad + bc)x + bd = 0

Then take the ratio of the ad/bc (in the brackets) - is it just luck that I got it right this way? Thanks for the help!

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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06 Aug 2016, 04:32
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AfricanPrincess wrote:
Hello, can someone please explain why this works:
ax(cx + d) = -b(cx + d)
axc^2 + axd = -bcx - bd
ax^2 + axd + bcx + bd = 0
ax^2 + (ad + bc)x + bd = 0

Then take the ratio of the ad/bc (in the brackets) - is it just luck that I got it right this way? Thanks for the help!

Posted from my mobile device

In the highlighted portion above, you have missed c in acx^2.

Also, in the bracket you have the sum of ad and bc not the product.

Now, if I consider your statement

acx^2 + (ad + bc)x + bd = 0

There is a very good solution provided by chetan2u in the forum. You can refer to that solution.

Else, you can simply do it in this way :

ax(cx + d) = -b(cx + d)

=> either cx+d=0 => x=-d/c ---(1)
or
ax=-b => x=-b/a ---(2)

Divide (1) by (2) or vice versa, you will get ad/bc or bc/ad.
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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03 Aug 2017, 13:29
This is a 'could be' question - don't overly complicate it.

$$ax(cx+d)=-b(cx+d) \rightarrow acx^2 +adx=-bcx-bd \rightarrow acx^2+x(ad+bc)+bd$$

Quadratics are easier to work with when the leading coefficient is $$1$$

Since this is a COULD BE question, then let's say $$a=c=1$$

Now we have

$$x^2+x(ad+bc)+bd$$

Since this is a COULD BE question with a well-formed quadratic, the roots could definitely be $$ad/bc$$, since $$ad$$ and $$bc$$ sum to the coefficient of the second term and multiply to the third term

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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18 Aug 2017, 18:23
one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation..
can anyone plz explain how to solve this equation
thank you

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]

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28 Dec 2017, 12:36
ax(cx+d)=-b(cx+d)

either cx+d = 0:
then, x = -d/c ..... (solution 1)

or cx + d not equal to 0:
then cancel it from both sides and;
ax = -b
i.e. x = -b/a ........ (solution 2)

therefore; ratio of 2 solutions = (-d/c) / (-b/a)

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation   [#permalink] 28 Dec 2017, 12:36
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