77th Problem 2016

There are two different solutions:-

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$\text {SOLUTION 1}$

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The nature of the roots can be found out by finding the discriminant which can in turn be found by applying the formula:- $b^2 - 4ac$

The equation can be written as $x^2 -6x+9=0$
Here, $a = 1 ; b = -6 ; c = 9$
$D = {(-6)}^2 - (4×1×9)$
$D = 36- (36)$
$D = 36-36$
$D = 0$ which is equal to zero.
So, the roots of the equation are real and equal.

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$\text {SOLUTION 2}$

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The equation can be written as $x^2 - 6x + 9 =0$
$x^2 - 3x -3x + 9= 0$
$x(x-3)-3(x-3) = 0$
$(x-3)(x-3) = 0$
$\therefore$ The roots are $3$ and $3$ . They are real and equal.

For the quadratic equation $Ax^2 + Bx + C = 0$ ,

if $B^2 = 4AC$ , the roots are equal

if $B^2 > 4AC$ , the roots are real and distinct

if $B^2 < 4AC$ , the roots are imaginary or non-real.

$x^2 - 6x + 6 = -3$ $\color{#D61F06}\implies$ $x^2 - 6x + 9 = 0$ $\color{#D61F06}\implies$ $\boxed{B^2 = 4AC}$ $\large\therefore$ The roots are equal.