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

Question

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}

KarishmaB · Accepted Answer

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

Their ratio could be bc/ad or ad/bc.

Answer (E)

chetan2u · Answer

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

ScottTargetTestPrep · Answer

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)

ad/bc

Answer: E

stonecold · Answer

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
we can see that the ratio may be ad/bc or bc/ad
Hence E

OptimusPrepJanielle · Answer

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?

DensetsuNo · Answer

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

AfricanPrincess · Answer

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

abhimahna · Answer

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.

spence11 · Answer

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

Answer E

karishmajain06 · Answer

one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation..
acx^2+x(ad+bc)+bd=0
can anyone plz explain how to solve this equation
thank you

KD1702 · Answer

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) 
 i.e. ratio = ad/bc ..... (answer)

Miracles86 · Answer

Excellent explanation!!!! I got to the point of (cx+d)(ax + b) = 0 but then started to multiply everything and ended up in a huge mess... Obviously that either cx+d=0 or ax+b=0.... that's a basic quadratic expression rule..

bp2013 · Answer

i don't mean to bump an old post, but i follow along your process all the way until the end where you get the ratio of ad/bc from (-d/c)/(-b/a). i'm not following. is there something basic i'm missing on why you're taking the denominator of one/numerator of the other, vis verse, and making them both positive?

push12345 · Answer

Rephrasing statement of question

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

Roots are -b/a or -d/c

So ratio would be bc/ad or ad/bc

Give kudos if it helps

Posted from my mobile device

CAMANISHPARMAR · Answer

The trick to answer this question correctly was to note that the equation is already written in the standard solution form. One can solve for the roots and take the fraction to get the correct answer.

BrentGMATPrepNow · Answer

GIVEN:  ax ( cx + d ) =  -b ( cx + d )
Add  b ( cx + d ) to both sides to get:  ax ( cx + d )  + b ( cx + d ) = 0
Rewrite as: ( ax + b )( cx + d ) = 0

This means either ( ax + b ) = 0 or ( cx + d ) = 0
Let's examine each case

Case a:  ax + b  = 0
So: ax = -b
Solve:  x = -b/a

Case b:  cx + d  = 0
So: cx = -d
Solve:  x = -d/c

Which of the following is a possible ratio of the 2 solutions?  
ASIDE: Why does the question ask for a POSSIBLE ratio? 
Well, the ratio of the solutions can be EITHER  (-b/a)  /  (-d/c)  OR  (-d/c)  /  (-b/a)

Let's check the first ratio. 
 (-b/a)  /  (-d/c)  = (-b/a)(c/-d) = bc/ad check the answer choices . . . not there. 
However, answer choice E is the reciprocal of bc/ad, so it must be the correct answer.

Not convinced? 
Let's confirm. 
 (-d/c)  /  (-b/a)  = (-d/c)(a/-b) = ad/bc = answer choice E

Cheers,
Brent

Kinshook · Answer

Given: Let a,b,c, and d be nonzero real numbers.

Asked: 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?

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

x = - b/a or -d/c

Ratio of solutions = bc/ad or ad/bc

IMO E

Basshead · Answer

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

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

ax + b = 0 
 x = \frac{-b}{a}

cx + d = 0 
 x = \frac{-d}{c}

= \frac{-b}{a }/ \frac{-d}{c }

= \frac{ad}{bc}

Answer is E.

rushabh27 · Answer

VeritasKarishma  Did you multiply the two equations?

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

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