7 - 6x- x2 = 0
so changing signs and reordering, x2 + 6x - 7 = 0
then x2 + 7x - x - 7 = 0
that is x(x + 7) - 1(x + 7) = 0
so that (x - 1)*(x + 7) = 0
which implies that x - 1 = 0 ie x = 1
or x + 7 = 0 ie x = -7
Chat with our AI personalities
None because without an equality sign the given expression is not an equation and so therefore no solutions are possible.
The quadratic equation is: Ax2+ Bx + C = 0-- The equation always has two solutions. They arex = 1/2A [ - B + sqrt(B2 - 4AC) ]x = 1/2A [ - B - sqrt(B2 - 4AC) ] .-- The solutions are real if ( B2 > or = 4AC ).-- The solutions are equal if ( B2 = 4AC ).-- The solutions are complex conjugates if ( B2 < 4AC ).-- The solutions are pure imaginary if ( B = 0 ) & (4AC>0 i.e. -4AC
A quadratic equation can have a maximum of 2 solutions. If the discriminant (b2-4ac) turns out to be less than 0, the equation will have no real roots. If the Discriminant is equal to 0, it will have equal roots. But, if the discriminant turns out to be more than 0,then the equation will have unequal and real roots.
Solutions: x = 9 and x = 1 Factored: (x-9(x-1) = 0 Equation: x2-10x+9 = 0
No. A pair of linear equation can have 0 solutions (they are parallel), or one solution (they cross at one point) or an infinite number of solutions (they represent the same line).