What is the nature of the roots of this equation:

${ x }^{ 2 }-6x+6=-3$

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Check out the set:
2016 Problems
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Not Real
Real and Equal
Real and Distinct

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There are two different solutions:-

$\text {SOLUTION 1}$

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.

$\text {SOLUTION 2}$

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.