Question

The quadratic equation ax2 – 4ax + 2a + 1 = 0 has repeated roots, if a =​

Answer 1

Understanding Concept :-

Let us consider a quadratic equation αx² + βx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

$\quad\red{ \underline { \boxed{ \sf{Discriminant, D = β² - 4αc}}}}$

Given Equation:-

$\red{\leadsto}\:$$\sf{ax^2 – 4ax + 2a + 1 = 0}$$\sf$

Solution :-

In the equation, $\:\:\sf{ax^2 – 4ax + 2a + 1 = 0}$

• α = a
• β = -4a
• c = 2a+1

Now,

$\quad\green{ \underline { \boxed{ \sf{Discriminant, D = β² - 4αc}}}}$

D = 0 for repeated roots

$\begin{gathered}\\\\\implies\quad \sf (-4a)^2-4\times a \times 2a+1=0\\\end{gathered}$

$\begin{gathered}\\\\\implies\quad \sf 16a^2 -8a^2-4a=0\\\end{gathered}$

$\begin{gathered}\\\\\implies\quad \sf 8a^2-4a=0\\\end{gathered}$

$\begin{gathered}\\\\\implies\quad \sf 4a(2a-1)=0\\\end{gathered}$

Thus,

$\begin{gathered}\\ \quad \maltese\:\:\boxed{\sf {4a =0 }}\quad and\: \quad \boxed{\sf{2a-1=0}} \\\end{gathered}$

$\begin{gathered}\\ \quad \maltese\:\:\boxed{\sf {a =0}}(rejected) \quad and\quad\: \boxed{\sf{2a=1}}\\\end{gathered}$

$\begin{gathered}\\ \quad\: \boxed{\sf { a=\frac{1}{2}}} \\\end{gathered}$

Therefore, $a=\dfrac{1}{2}$ for repeated roots.