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			 Quadratic functions 
			A quadratic function has the general form 
			  
			
			  
			The quadratic function is also known as a second degree polynomial. 
			 
			1. Solving quadratic equation 
			There are three methods of solving quadratic equations. 
			 
			2. Factorization 
			When the factors are multiplied that is expanded, you get back the quadratic equation.  
			 
			Example 
			  
			
				
					
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						 \(x^2+5x+6=0 \) 
						\(x+3x+2=0\) 
						\(x=-3 \text { and } x=-2\) 
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			Example 
			  
			
				
					
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						 \(2x^2-x-1=0 \) 
						\(x-12x+1=0 \) 
						\(x=-12 \text { and } x=1 \) 
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			3. Completing the square 
			For the quadratic expression   
			  
			
			  
			you complete the square by first factoring out a : 
			  
			
				
					
						| \(a(x^2+{b \over a} \times x +{c \over a})\) | 
					 
				
			 
			  
			and the using the formula : 
			  
			
				
					
						| \(a[(x+{b \over 2a}^2)+{c \over a}-{b^2 \over 4a^2}]\) | 
					 
				
			 
			  
			Example 
			For 
			  
			
			  
			the coefficients are a=1, b=5 and c=6. Then, using the above formula, you have 
			  
			
				
					
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						 \((x+{5 \over 2(1)})^2+6-{5^2 \over 4(1)^2}=0\) 
						\((x+{5 \over 2})^2+6-{25 \over 4}=0\) 
						\( (x+{5 \over2 })^2-{1 \over 4}=0\) 
						\((x+{5 \over 2})^2={1 \over 4}\) 
						\(x+{5 \over 2}-{1 \over 2} \text { and } x+{5 \over 2}={1 \over 2}\) 
						\(x=-{1 \over 2}-{5 \over 2} \text{ and } x={1 \over 2}-{5 \over 2}\) 
						\(x=-3 \text{ and } x=-2 \) 
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			Example 
			For 
			  
			
			  
			the coefficients are a=2, b=-1 and c= -1. Then, using the  completing square method, you have 
			  
			
				
					
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						 \( 2(x^2-{1 \over 2}x-{1 \over 2})=0\) 
						\( 2[(x-{1 \over 4})^2-{1 \over 2}-{1 \over 16}]=0\) 
						\( 2[(x-{1 \over 4})^2-{9 \over 16}]=0\) 
						\( (x-{1 \over 4})^2-{9 \over 16}=0\) 
						\( (x-{1 \over 4})^2={9 \over 16}\) 
						\( x-{1 \over 4}=-{ 3 \over 4} \text { and } x-{1 \over 4}={3 \over 4} \) 
						\( x=-{3 \over 4}+{1 \over 4} \text { and } x={3 \over 4} +{1 \over 4}\) 
						\(x=-{1 \over 2} \text { and } x=1 \) 
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			4. Quadratic formula 
			For the quadratic equation   
			  
			
			  
			the quadratic formula to find x is : 
			  
			
				
					
						| \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) | 
					 
				
			 
			  
			Example 
			For 
			  
			
			  
			where you have seen that the coefficients are a=1, b=5 and c=6, 
			  
			
				
					
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						 \(x = {-5 \pm \sqrt{5^2-4(1)(6) }\over 2(1)}\) 
						\(x = {-5 \pm \sqrt{25-24} \over 2}\) 
						\(x = {- 5\pm \sqrt{1} \over 2}\) 
						\(=-{5-1 \over 2}, {-5+1 \over 2}\) 
						\(={-6 \over 2}, {-4 \over 2}\) 
						\(=-3,-2 \) 
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			5. Roots of quadratic equation 
			The number of roots and the type of roots of the quadratic equation 
			  
			
			  
			can be determined by the discriminant D=b2-4ac. 
			
				- If D>0, the quadratic equation has two different real roots
 
				- If D=0, the quadratic equation has one real root, repeated twice
 
				- If D<0, the quadratic equation has two different complex roots
 
			 
			 
			Example 
			Notice that in  
			  
			
			  
			given above,  
			  
			
			  
			and so you had two different real roots x=-3 and x=-2. 
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