The quadratic equations x^2 − 6x + a = 0 and x^2 − cx + 6 = 0 have one

Good Question!

Let the common root = \(m\) & the other roots be \(4k\) & \(3k\) respectively
--> The roots of 1st equation are (\(m, 4k\)) and the roots of the 2nd equation are (\(m, 3k\))

Formula: For the quadratic equation \(ax^2 + bx + c = 0\), Sum of the roots = \(\frac{-b}{a}\) & Product of the roots = \(\frac{c}{a}\)

For \(x^2 − 6x + a = 0\),
Sum of the roots = \(m + 4k\) = \(\frac{-(-6)}{1} = 6\)
--> \(m + 4k = 6\) ........ (1)

and for \(x^2 − cx + 6 = 0\)
Product of the roots = \(m*3k\) = \(\frac{6}{1}\)
--> \(m*k = 2\)
--> \(k = \frac{2}{m}\) ........ (2)

Substituting (2) in (1),
--> \(m + 4(2/m) = 6\)
--> \(m^2 + 8 = 6m\)
--> \(m^2 - 6m + 8 = 0\)
--> \((m - 2)(m - 4) = 0\)
--> \(m = 2\) or \(4\)

Given that, all roots should be integers
Case 1: If \(m = 2, k = \frac{2}{m} = 1\), So, other roots are \(4k\) & \(3k\) = \(4\) & \(3\) respectively --> Possible
Case 2: If \(m = 4, k = \frac{2}{m} = \frac{1}{2}\), So, other roots are \(4k\) & \(3k\) = \(2\) & \(3/2\) respectively --> Not Possible

So, \(m = 2\) ONLY

Option B

Let \(m\) be the common root
And other roots —\(4k\) and \(3k \)
(1st) —> \(m*4k = a\)
(2nd)—> \(m* 3k = 6\)
—> \(\frac{a}{4} = 2\) —>\( a= 8 \)
\(x^{2} —6x + a = 0\) —> \(x^{2} —6x +8=0 \)
\((x—2)(x—4) = 0 \)

Well, in order to find the common root, just look at the second equation : \(x^{2} —cx+ 6 =0\)
—> since roots are integers, \(6\) could the product of the \(1\) and \(6\) or the product of \(2\) and \(3\).
—> we have \(2\) as the common root.

Answer(B)

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Let the common root of the quadratic equations be A. The question says that the other roots of the two equations are in the ration of 4:3; therefore, we can take these roots as 4k and 3k.

The standard form of a quadratic equation is a\(x^2\)+bx+c = 0. Let’s compare the two equations with the standard form:

\(x^2\) -6x + a = 0. The value of a = 1, b = -6 and c =a.

\(x^2\) – cx + 6 = 0. The value of a = 1, b = - c and c = 6.

The sum of the roots of any quadratic equation is given by –(\(\frac{b}{a}\)) and the product of the roots is given by (\(\frac{c}{a}\)).

For \(x^2\)- 6x + a = 0, sum of roots = 6 and product of roots = a.

Substituting the values of the roots, we have
A+4k = 6 and A*4k = a.

For \(x^2\) – cx + 6 = 0, sum of roots = c and product of roots = 6.

Substituting the values of the roots, we have
A+3k = c and A*3k = 6.

Simplifying, we get Ak = 2 or A = \(\frac{2}{k}\).

Substituting this in A+4k = 6, we have \(\frac{2}{k}\) + 4k = 6. Rearranging terms, we have 4\(k^2\) -6k + 2 = 0 OR 2\(k^2\) – 3k + 1 = 0. The roots of this equation are k = 1 or k = ½. k = ½ does not give 3k as an integer and hence can be ruled out.

Therefore, A = \(\frac{2}{k}\) = \(\frac{2}{1}\) = 2.

The correct answer option is B.

In questions on quadratic equations,when the value of one root is given and the relationship between the roots is given, it's a good idea to deal with the sum and the product of the roots instead of trying to solve the equation. Because one root is given and you know how this root is related to the other one, you will generally be able to develop equations for the sum and product of roots and hence be able to solve for the unknown roots.

Hope that helps!

See the attachment.

Let \(m =\) the common root.

Since the other roots are in a ratio of 4:3, it seems likely that the equations can be factored as follows:
\(x^2 − 6x + a = (x-4)(x-m)\)
\(x^2 − cx + 6 = (x-3)(x-m)\)

To yield \(-6x\) in the first equation and \(+6\) in the second equation, \(m=2\):
\(x^2 − 6x + a = (x-4)(x-2)\)
\(x^2 − cx + 6 = (x-3)(x-2)\)


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I know this is a silly question, but will these kinds of question come on the GMAT ?

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