x=(-b+-sqrt(b^(2)-4ac))/(2a)

Sohan and Mohan are playing a chess on Sunday. The chess board contains equal squares and the are of each equal square is 6.25 m^(2) . A border round the board is 2 cm wide. For equation ax^(2)+bx+c=0, x = (-b pm sqrt(b^(2)-4ac))/(2a) is called the.......

Theorem : The roots of ax^(2)+bx+c=0 are (-b pm sqrt(b^(2)-4ac))/(2a)

The roots of the equation ax^(2)+bx+c=0 where a!=0 ,are: 1) (b+-sqrt(b^(2)-4ac))/(2a), 2) (-b+-sqrt(b^(2)-4ac))/(2c), 3) (-b+sqrt(b^(2)-4ac))/(2a), 4) (2a)/(2a-b+sqrt(b^(2)-4ac)

The quadratic formula states that x=(-bsqrt(b^2-4ac))/(2a) . Find x given that a=2, b=4 and c=2

The roots of the quadratic equation ax^(2) + bx + c = 0 (a ! = 0 ) are given x = (-b pm sqrt(b^(2) - 4ac))/(2a) are (i) real and distinct roots if D gt 0 (ii) repeated roots if D = 0 no real roots if D lt 0 , where D = b^(2) - 4ac The nature of the roots of quadratic equation 4x^(2) + 20x + 25 = 0 is

Four steps to derive the quadratic formula are shown below . (I) x^(2)+(bx)/a=(-c)/a (II) (x+b/(2a))^(2)=(b^(2)-4ac)/(4a^(2)) (III) x =pm sqrt((b^(2)-4ac)/(4a^(2)))-b/(2a) (IV) x^(2)+(bx)/a +(b/(2a))^(2)=(-c)/a + (b/(2a))^(2) What is the correct order for these steps ?

The roots of the equation ax^2 - bx + c = 0 are .. 1.x =' (-b pm sqrtb^2 - 4ac)/(2a) 2.x =' (b pm sqrtb^2 - 4ac)/(2a)'

if alpha , beta are root of ax^2+bx+c=0 then (1/alpha^2+1/beta^2)^2 (a) (b^(2)(b^(2)-4ac))/(c^(2)a^(2)) (b) (b^(2)(b^(2)-4ac))/(ca^(3)) (c) (b^(2)(b^(2)-4ac))/(a^(4)) (d) (b^(2)-2ac)^2/(c^(4))

If (sqrt(a + 2b) + sqrt(a - 2b))/(sqrt(a + 2b) - sqrt(a - 2b)) = sqrt3 and a^(2) + b^(2) = 1 , then the find values of a and b. (b) If sqrt((x - sqrt(a^(2) - b^(2)))^(2) + y^(2)) + sqrt((x + sqrt(a^(2) - b^(2)))^(2) + y^(2)) = 2a then prove that (x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1

If b^2 - 4ac