How do you solve # x^2-4x-7=0#?

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Answer 1 · 2016-06-04T13:50:04

#x = 5.317 or x = -1.317#

This is a quadratic equation. There are 3 methods for solving them:
Factorise, completing the square or using the quadratic formula.

The given equation does not factorise, so option 1 falls away.

My choice would be completing the square because:
the coefficient of #x^2# is 1, and the coefficient of the middle term is even,

Student often think completing the square is difficult, but in a case like this it is quicker and easier than using the formula.

#x^2 - 4x -7 = 0#
Move the constant to the RHS: # x^2 - 4x " " = 7#

Add on the square of, half of the middle coefficient to both sides:
#( 4÷ 2 = 2 and 2^2 = 4)#

# x^2 - 4x + color(red)4 = 7 + color(red)4#

The LHS is now the square of a binomial

#(x - 2)^2 = 11 " "# Now find the square root of each side

#x - 2 = +-sqrt11 " "# Solve for x twice.

#x = sqrt11 + 2 " " or x = -sqrt11 + 2#

#x = 5.317 " " or x = -1.317#