Question

The quadratic equations \[{x^2} - 6x + a = 0\] and \[{x^2} - cx + 6 = 0\] have one root in common. The other roots of the first and second equations are integers in the ratio \[4:3\] . Then the common root is
1. \[2\]
2. \[1\]
3.\[4\]
4. \[3\]

Answer 1

Hint: In order to solve for the common root initiate with finding the sum of the roots and the products of roots for both the quadratic equation, using the formula: Sum of the roots $ = - \dfrac{b}{a}$ and Product of roots $ = \dfrac{c}{a}$ for the quadratic equation \[a{x^2} + bx + c = 0\]. Compare the roots of both equations and find out the common root.

Complete step-by-step answer:
We are given two quadratic equations \[{x^2} - 6x + a = 0\] and \[{x^2} - cx + 6 = 0\] that have one root in common.
As, we know that any quadratic equation in the variable \[x\] is of the form \[a{x^2} + bx + c = 0\].
And, we know Sum of the roots $ = - \dfrac{b}{a} $ and Product of roots $ = \dfrac{c}{a} $.
Let \[\alpha \] and \[4\beta \] be roots of \[{x^2} - 6x + a = 0\] and, \[\alpha \] and \[3\beta \] be the roots of \[{x^2} - cx + 6 = 0\] where $\alpha $ is the common root, then by using the above mentioned properties we have ,
For \[{x^2} - 6x + a = 0\]:
\[\alpha + 4\beta = 6\] (Sum of roots) … (1)
\[4\alpha \beta = a\] (Product of roots) …. (2)
Similarly, for \[{x^2} - cx + 6 = 0\]
\[\alpha + 3\beta = c\] (Sum of roots) … (3)
\[3\alpha \beta = 6\] (Product of roots) …. (4)
On dividing equation 2 by 4, we get:
$\dfrac{{4\alpha \beta }}{{3\alpha \beta }} = \dfrac{a}{6}$
$ \Rightarrow \dfrac{4}{3} = \dfrac{a}{6}$
Multiplying both sides by 6:
$
   \Rightarrow \dfrac{4}{3} \times 6 = \dfrac{a}{6} \times 6 \\
   \Rightarrow \dfrac{4}{3} \times 6 = a \\
   \Rightarrow 8 = a \;
 $
Therefore, we get \[a = 8\] .
Now putting this value in equation (2) we get \[\alpha \beta = 2\].
Therefore, our first equation becomes \[{x^2} - 6x + 8 = 0\]
Roots of this equation using the quadratic formula, are
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
= \dfrac{{6 \pm \sqrt {{6^2} - 4(8)} }}{2} \\
= \dfrac{{6 \pm \sqrt {36 - 32} }}{2} \\
= \dfrac{{6 \pm \sqrt 4 }}{2} \\
= \dfrac{{6 \pm 2}}{2} = 3 \pm 1\;
\]
Therefore, we get
\[x = \left( {3 + 1} \right),\left( {3 - 1} \right)\]
\[x = 4{\text{ , 2}}\]
 \[x = 2,4\]
Now, if \[\alpha = 2\] and \[4\beta = 4\] then we get \[3\beta = 3\].
If \[\alpha = 4\] and \[4\beta = 2\] then we have \[3\beta = \dfrac{3}{2}\] (not an integer).
Therefore, common root is \[x = \alpha = 2\]
So, the correct answer is “Option 1”.

Note: Remember, it’s not important to take \[4\beta {\text{ and 3}}\beta \] as the second root, we could have taken any other variable, but since, we were given that the ratio of the roots is \[4:3\] that’s why we took the respective value in order to get the given ratio to solve easily.
Roots for a quadratic equation can be found using two methods, first is the mid-term factorization by splitting the mid-term, and second method is using the quadratic formula.