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
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).
0 = 0 is an identity and not an equation. Equations have solutions, identities do not.
If the discriminant of a quadratic equation is less then 0 then it will have no real solutions.
x2 + 49 = 0
In general, you cannot: it all depends on the domain.x + 2 = 0 has no solutions is the set of positive integers but does have one if the domain is the integers.2x - 3 = 0 has no solutions if the domain is integers, but there is one solution if the domain is the rationals.x2 - 2 = 0 has no solutions if the domain is the rationals but there are two solutions if the domain is the reals.x2 + 2 = 0 has no solutions if the domain is the reals but there are two solutions if the domain is the complex numbers.Cos(x) = 1 has no solutions if the domain is (0, 360) but two solutions for the domain [0, 360].
The solutions of the equation (if any) remain unchanged.
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.
It has infinitely many solutions.
If the discriminant of b2-4ac of the quadratic equation is greater the 0 then it will have 2 solutions.
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).