Question

What is the least positive value of \[a\] for which the equation \[2{x^2} + \left( {a - 10} \right)x + \dfrac{{33}}{2} = 2a\] has real roots?

Answer 1

Hint: First we rewrite the given equation in the form \[a{x^2} + bx + c = 0\] and then compare the equation with \[a{x^2} + bx + c = 0\]. Then we will apply the discriminant formula \[{b^2} - 4ac\]. For real root we apply the condition \[{b^2} - 4ac \ge 0\]. Then simply the inequality to get the value of \[a\].

Formula Used:
The discriminant formula for the quadratic equation \[a{x^2} + bx + c = 0\] with real roots is \[{b^2} - 4ac \ge 0\].

Complete Step-by Step Solution:
Given equation is \[2{x^2} + \left( {a - 10} \right)x + \dfrac{{33}}{2} = 2a\],
We will take all terms to one side, we will get,
\[2{x^2} + \left( {a - 10} \right)x + \dfrac{{33}}{2} - 2a = 0\]
We will compare the equation with the quadratic equation \[a{x^2} + bx + c = 0\],
So, here
\[a = 2\],\[b = a - 10\], and \[c = \dfrac{{33}}{2} - 2a\],
Now we will substitute the values of \[a\],\[b\] and \[c\], in \[{b^2} - 4ac \ge 0\].
\[ \Rightarrow {\left( {a - 10} \right)^2} - 4\left( 2 \right)\left( {\dfrac{{33}}{2} - 2a} \right) \ge 0\]
Now we will simplify the equation, we will get,
\[ \Rightarrow {\left( {a - 10} \right)^2} - 4\left( {33 - 4a} \right) \ge 0\]
Now we will further simplify by expanding the first term,
\[ \Rightarrow {a^2} - 20a + 100 - 132 + 16a \ge 0\]
Now we will further simplify, we will get,
\[ \Rightarrow {a^2} - 4a - 32 \ge 0\]
Now we will further simplify by factoring the given equation, we will now get,
\[ \Rightarrow {a^2} - 8a + 4a - 32 \ge 0\]
Now we will take out the common factor, we will get,
\[ \Rightarrow a\left( {a - 8} \right) + 4\left( {a - 8} \right) \ge 0\],
Now we will again take out the common factor, we will get,
\[ \Rightarrow \left( {a + 4} \right)\left( {a - 8} \right) \ge 0\]
Now we will equate each term to \[0\],
\[ \Rightarrow a + 4 \ge 0\], and \[ \Rightarrow a - 8 \ge 0\]
Now we will get the solution set as
\[ \Rightarrow a \in \left( { - \infty , - 4} \right] \cup \left[ {8,\infty } \right)\]
The least possible positive value of \[a\] is \[8\].

  Note: Students often make a common mistake to solve the inequality. They take \[\left( {a + 4} \right)\left( {a - 8} \right) \ge 0\]\[ \Rightarrow a + 4 \ge 0\] or, \[ \Rightarrow a - 8 \ge 0\]. So the solution becomes \[\left[ { - 4,\infty } \right)\] and the least possible positive value of \[a\] becomes \[1\].