Find the least a, for which two equations have a common root
$2 x^{2} - 3 x + 1 = 0$
and
$2 x^{2} - (a + 3) x + 3 a = 0$

Question

Find the least a, for which two equations have a common root  2x2−3x+1=0  and  2x2−(a+3)x+3a=0

Karli Kaiser · Accepted Answer

Step 1
I think you can just do it in a concrete way. The roots of

2 x^{2} - 3 x + 1 = 0

are

\frac{1}{2}

and 1. By quadratic formula, the roots of

2 x^{2} - (a + 3) x + 3 a = 0

are

\frac{a + 3 - \sqrt{a^{2} - 18 a + 9}}{8}

and

\frac{a + 3 + \sqrt{a^{2} - 18 a + 9}}{8}

If they have common roots, we have the following possibilities:

\frac{a + 3 - \sqrt{a^{2} - 18 a + 9}}{8} = \frac{1}{2}

or

\frac{a + 3 - \sqrt{a^{2} - 18 a + 9}}{8} = 1

or

\frac{a + 3 + \sqrt{a^{2} - 18 + 9}}{8} = \frac{1}{2}

or

\frac{a + 3 + \sqrt{a^{2} - 18 a + 9}}{8} = 1

Each one of them is a quadratic equation in a which can be solved.

pripravyf · Accepted Answer

Step 1  The resultant of the two polynomials is  [2−31002−312−a−33a002−a−33a]  Equating this determinant to 0 we get  2(2a−1)(5a−2)=0  giving  a=12 and a=25

RizerMix · Accepted Answer

No derivative knowledge is needed. A common zero of your two polynomials is a root of their difference. So it must be a root of the equation ax=3a−1. It is easy to see that a≠0. So any common root must be equal to 3−1a. Substitute in the first equation and solve for a. Because this is homework, we omit the rest of the calculation. But after a while you should get a quadratic in a. Comment: For various reasons, it is nice to put off dividing as long as possible. Since a≠0, we can rewrite the first equation as a2x2−3a2x+a2=0. Then we can substitute 3a−1 for ax. This yields 2(3a−1)2−3a(3a−1)+a2=0, and then simplification is pleasant and quick.

Find the least a, for which two equations have a

Answered question

Answer & Explanation

New Questions in Algebra I