The quadratic equation ax^2 + bx + c = 0 satisfies a - 2b + 4c = 0 and has two equal real roots. Which conclusion is correct: a = b, a = c, b = c, or c = 4a?

Step-by-step solution

1. 1

   Translate 'two equal real roots' into algebra. A quadratic has a repeated root exactly when its discriminant vanishes:

   $\Delta=b^2-4ac=0\ \Longrightarrow\ b^2=4ac$

   (Implicitly $a\neq 0$, otherwise the equation is not quadratic.)

2. 2

   Rearrange the given linear relation to isolate $b$. From $a-2b+4c=0$:

   $2b=a+4c\ \Longrightarrow\ b=\frac{a+4c}{2}$

   The factor of $2$ in front of $b$ is a deliberate hint that $b$ is meant to be eliminated.

3. 3

   Substitute into the discriminant condition.

   $\left(\frac{a+4c}{2}\right)^2=4ac$

   Multiply both sides by $4$:

   $(a+4c)^2=16ac$

4. 4

   Expand and recognise a perfect square.

   $a^2+8ac+16c^2=16ac$

   $a^2-8ac+16c^2=0$

   $(a-4c)^2=0\ \Longrightarrow\ a=4c$

5. 5

   Push one step further to reach an offered option. The relation $a=4c$ is not among the choices, so substitute it back into $2b=a+4c$:

   $2b=4c+4c=8c\ \Longrightarrow\ b=4c$

   Since $a=4c$ and $b=4c$, we get $a=b$ — option (A).

6. 6

   Verify with a concrete example. Take $c=1$, so $a=b=4$: the equation is $4x^2+4x+1=0$. Its discriminant is $4^2-4(4)(1)=16-16=0$ $\\checkmark$ (repeated root $x=-\\tfrac12$), and the side condition gives $4-2(4)+4(1)=4-8+4=0$ $\\checkmark$. Note that $c=4a$ (option D) would read $1=16$, which is false — the correct relation runs the other way, $a=4c$.