Maths-
General
Easy

Question

Find the solution of the system of equations.
y equals x squared minus 5 x minus 8
y equals negative 2 x minus 4

hintHint:

A quadratic equation of the form  x2 +( a +b) x + ab can be simplified into (x + a) (x + b). This can be further solved by equating it to zero.
We are asked to solve the set of equation.

The correct answer is: We are asked to solve the set of equation.


    Step 1 of 3 :
    The LHS of the given set of equation has the same value. Equate the RHS to down the set to a single variable.
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell x squared minus 5 x minus 8 equals negative 2 x minus 4 space space space end cell row cell x squared minus 5 x plus 2 x minus 8 plus 4 equals 0 end cell row cell x squared minus 3 x minus 4 equals 0 space space space space space space space space space space space space space space end cell end table
    Step 2 of 3:
    Solve the simplified equation:
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell x squared minus 3 x minus 4 equals 0 space space space space space space space space space space space end cell row cell x squared minus 4 x plus x minus 4 equals 0 space space space space space space end cell row cell x left parenthesis x minus 4 right parenthesis plus 1 left parenthesis x minus 4 right parenthesis equals 0 end cell row cell left parenthesis x plus 1 right parenthesis left parenthesis x minus 4 right parenthesis equals 0 space space space space space space space space space end cell row cell left parenthesis x plus 1 right parenthesis equals 0 straight & left parenthesis x minus 4 right parenthesis equals 0 end cell row cell x equals negative 1 straight & x equals 4 space space space space space space space space space space space space space end cell end table
    Step 3 of 3:
    Substitute the values of  in any of the equation to get the values of y.
    When x = -1,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell y equals negative 2 x minus 4 space space space space space space space space space space end cell row cell y equals negative 2 left parenthesis negative 1 right parenthesis minus 4 space space space space space end cell row cell y equals 2 minus 4 space space space space space space space space space space space space space space space space space end cell row cell y equals negative 2 space space space space space space space space space space space space space space space space space space space end cell end table
    when x = 4,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell y equals negative 2 x minus 4 space space end cell row cell y equals negative 2 left parenthesis 4 right parenthesis minus 4 end cell row cell y equals negative 8 minus 4 space space space space space end cell row cell y equals negative 12 space space space space space space space space space end cell end table
    Thus, the required sets of  solutions are ( -1, -2) and ( 4, -12).

    The maximum number of solutions a polynomial of degree two could have is two.

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