Solve for x: 1.1.1 (x−3)(x+1)=0 1.1.2 √x=512 1.1.3 x(x−4)<0 Given: f(x)=x²−5x+2 1.2.1 Solve for x if f(x)=0 1.2.2 For which values of c will f(x)=c have no real roots? Solve for x and y: x=2y+2 x²−2xy+3y²=4 1.4 Calculate the maximum value of S if S=6/(x²+2) - NSC Mathematics - Question 1 - 2017 - Paper 1

To solve the equation $\sqrt{x} = 512$, we first square both sides:

[ x = 512^2 = 262144 ] So, the solution for x is ( x = 262144 ).

To solve the inequality $x(x−4)<0$, we need to find the critical points. Setting the equation equal to zero:

1. Critical Points: [ x(x−4) = 0 ]
   This gives us ( x = 0 ) and ( x = 4 ).

2. Test Intervals: We test the intervals formed by these critical points: ((-∞, 0), (0, 4), (4, ∞)).

   • For ((-∞, 0)): Choose ( x = -1 ) → ( (-1)(-5) > 0 )
   • For ((0, 4)): Choose ( x = 2 ) → ( (2)(-2) < 0 )
   • For ((4, ∞)): Choose ( x = 5 ) → ( (5)(1) > 0 )

Thus, the solution to the inequality is ( 0 < x < 4 ).

To find the roots of the function $f(x)=x^2−5x+2$, we set it equal to zero: [ x^2 - 5x + 2 = 0 ] We can use the quadratic formula: [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ] where ( a=1, b=-5, c=2 ).

Calculating the discriminant: [ b^2 - 4ac = (-5)^2 - 4(1)(2) = 25 - 8 = 17 ]

Thus, [ x = \frac{5 \pm \sqrt{17}}{2} ] This gives us two solutions: ( x = \frac{5 + \sqrt{17}}{2} ) and ( x = \frac{5 - \sqrt{17}}{2} ).

For the equation ( f(x)=c ) to have no real roots, the discriminant must be less than zero:

Starting from the original function ( f(x)=x^2−5x+(2-c) ), we identify:\n- ( a = 1 ), ( b = -5 ), ( c = 2 - c )
The discriminant is evaluated as: [ b^2 - 4ac = (-5)^2 - 4(1)(2-c) = 25 - (8 - 4c) = 25 + 4c - 8 = 4c + 17 ]

For no real roots, we set this less than zero: [ 4c + 17 < 0 ] [ 4c < -17 ] [ c < -\frac{17}{4} ] Thus, for values of ( c < -4.25 ), the equation has no real roots.

To solve the system, we substitute ( x = 2y + 2 ) into the second equation:

1. Substitute:
   [ (2y + 2)^2 - 2(2y + 2)y + 3y^2 = 4 ]
2. Expanding gives:
   [ 4y^2 + 8y + 4 - 4y^2 - 4y - 6y^2 = 4 ] Simplifying leads to:
   [ -6y^2 + 4y + 4 - 4 = 0 ]
3. Rearranging gives:
   [ -6y^2 + 4y = 0 ] Factoring:
   [ 2y(3y - 2) = 0 ] Thus, ( y = 0 ) or ( y = \frac{2}{3} ).
4. For ( y = 0 ):
   ( x = 2(0) + 2 = 2 )
   For ( y = \frac{2}{3} ):
   ( x = 2(\frac{2}{3}) + 2 = \frac{4}{3} + 2 = \frac{10}{3} ).

Thus, the pairs ( (2, 0) ) and ( (\frac{10}{3}, \frac{2}{3}) ) are solutions.

To calculate the maximum value of
$S = \frac{6}{x^2 + 2}$, we note that the expression is minimized when ( x^2 ) is maximized. Since the denominator is always positive, the maximum occurs at the minimum of ( x^2 + 2 ).

The minimum value of ( x^2 ) is 0, when ( x = 0 ) thus: [ S_{max} = \frac{6}{0 + 2} = 3 ] Therefore, the maximum value of S is 3.