Question

The least possible value of $a$ for which the equation,$2x^2+\left(a-10\right)x+\frac{33}{2}=2a$ has real root is

Answer 1

Solve the given expression to find the least possible value of the constant

Given expression ,$2x^2+(a-10)x+\frac{33}{2}=2a$

$\Rightarrow 2x^2+(a-10)x+\frac{33}{2}-2a=0$

Now, comparing above equation with $a{x}^2+bx+c=0$

$a=2,b=(a-10)$ and $c=\frac{33}{2}-2a$

For real roots ,we have a condition

$$\begin{gathered} D\ge 0 \\ \Rightarrow b^2-4ac\ge 0 \\ \Rightarrow {(a-10)}^2-4\left(2\right)(\frac{33}{2}-2a)\ge 0 \\ \Rightarrow {(a-10)}^2-4(33-4a)\ge 0 \\ \Rightarrow a^2-4a-32\ge 0 \\ \Rightarrow (a-8)(a+4)\ge 0 \end{gathered}$$

Now from set theory, we can say that

$a\in (-\infty ,-4]\cup [8,\infty )$

So least possible value will be $8$.

Therefore, the least possible value of $a$ is $8$.