Question

Find the real roots of the equation $x^2+2ax+\frac{1}{16}=-a+\sqrt{(a^2+x-\frac{1}{16})}$
where $0<a<\frac{1}{4}$.

• A. $x_1=\frac{(1-2a)}{2}+\sqrt{{\left(\frac{1-2a}{2}\right)}^2-\frac{1}{16}}$
• B. $x_2=\left(\frac{1-2a}{2}\right)-\sqrt{{\left(\frac{1-2a}{2}\right)}^2-\frac{1}{16}}$
• C. $x_2=\left(\frac{1+2a}{2}\right)-\sqrt{{\left(\frac{1+2a}{2}\right)}^2-\frac{1}{16}}$
• D. $x_1=\frac{(1+2a)}{2}+\sqrt{{\left(\frac{1+2a}{2}\right)}^2-\frac{1}{16}}$

Answer 1

Answer: (A), (B)

The correct options are
A $x_1=\frac{(1-2a)}{2}+\sqrt{{\left(\frac{1-2a}{2}\right)}^2-\frac{1}{16}}$
B $x_2=\left(\frac{1-2a}{2}\right)-\sqrt{{\left(\frac{1-2a}{2}\right)}^2-\frac{1}{16}}$

We have,
$x^2+2ax+\frac{1}{16}=-a+\sqrt{(a^2+x-\frac{1}{16})}$
Let $y=x^2+2ax+\frac{1}{16}$ ......(1)
and $y_1=-a+\sqrt{(a^2+x-\frac{1}{16})}$ ......(2)
$\therefore y=y_1$ .....(3)
From (2),
$y_1+a=\sqrt{(a^2+x-\frac{1}{16})}$
$\Rightarrow x=y_1^2+2a{y}_1+\frac{1}{16}x=y^2+2ay+\frac{1}{16}$ ......(4)
{From (3)}
Equation (1) and (4) represents parabolas, both parabolas symmetrical about the line
y=x .....(5)
From (1)& (5) we get,
$\Rightarrow x=x^2+2ax+\frac{1}{16}\Rightarrow x^2-(1-2a)x+\frac{1}{16}=0$
$\therefore x=\frac{(1-2a)\pm \sqrt{{(1-2a)}^2-\frac{1}{4}}}{2}$
For real roots ${(1-2a)}^2-\frac{1}{4}\ge 0$
$\therefore 1-2a>\frac{1}{2}\Rightarrow a<\frac{1}{4}$
Hence, $x_1=\frac{(1-2a)}{2}+\sqrt{{\left(\frac{1-2a}{2}\right)}^2\frac{1}{16}}and{x}_2=\left(\frac{1-2a}{2}\right)-\sqrt{{\left(\frac{1-2a}{2}\right)}^2\frac{1}{16}}$
are real roots and given fact satisfy the original equation.