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  • Which of the following equations has the sum of its roots as 3? (A)2x^{2}-3x+6=0\\ (B)-x^{2}+3x-3=0\\ (C)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0\\ (D)3x^{2}-3x+3=0

Which of the following equations has the sum of its roots as 3?          

\\(A)2x^{2}-3x+6=0\\ (B)-x^{2}+3x-3=0\\ (C)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0\\ (D)3x^{2}-3x+3=0  

Answers (1)

 (B) (B)-x^{2}+3x-3=0              

If (B)ax^{2}+bx+c=0  is a quadratic equation the sum of its roots are -\frac{b}{a}

(A)   
        2x^{2}-3x+6=0           

         Here b=-3,a=2

            Sum of its roots =-\frac{b}{a} = -\left ( \frac{-3}{2} \right )=\frac{3}{2}

(B)      

            -x^{2}+3x-3=0

            Here b=-3,a=3

            Sum of its roots =-\frac{b}{a}=\frac{-3}{-1}=3

(C)       

            \sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0   

            Here b=\frac{-3}{\sqrt{2}}, a=\sqrt{2}

            Sum of its roots = -\frac{b}{a}=-\left (\frac{-3}{\sqrt{2}} \right )\times \sqrt{2}=\frac{3}{2}

(D)     

            3x^{2}-3x+3=0

            Here b=-3,a=3

            Sum of its roots =-\frac{b}{a}=\frac{(-3)}{3}=\frac{3}{3}=1

            Hence only B has sum of its roots 3

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