Answers>Further Mathematics>A Level>Article

Solve x^2+8x-5=0 using completing the square

by completing the square we write the equation as (x+b/2)^2-b/2^2+c, in this case b=8 (the coefficient of x) and c=5 so we have (x+4)^2-16-5=0, which equals (x+4)^2-21=0. Now by rearranging we get (x+4)^2=21, which goes to x+4=+or-sqrt(21). Therefore x=sgrt21 -4 or x=-sqrt21 -4

Answered by Lucy H. • Further Mathematics tutor

3018 Views

See similar Further Mathematics A Level tutors

Solve x^2+8x-5=0 using completing the square

Related Further Mathematics A Level answers

Find the general solution to: d^(2)x/dt^(2) + 7 dx/dt + 12x = 2e^(-t)

Further Maths: How do you find the inverse of a 2 x 2 matrix?

Integrate cos(4x)sin(x)

The infinite series C and S are defined C = acos(x) + a^2cos(2x) + a^3cos(3x) + ..., and S = asin(x) + a^2sin(2x) + a^3sin(3x) + ... where a is a real number and |a| < 1. By considering C+iS, show that S = asin(x)/(1 - 2acos(x) + a^2), and find C.

Solve x^2+8x-5=0 using completing the square

Related Further Mathematics A Level answers

Find the general solution to: d^(2)x/dt^(2) + 7 dx/dt + 12x = 2e^(-t)

Further Maths: How do you find the inverse of a 2 x 2 matrix?

Integrate cos(4x)sin(x)

The infinite series C and S are defined C = a*cos(x) + a^2*cos(2x) + a^3*cos(3x) + ..., and S = a*sin(x) + a^2*sin(2x) + a^3*sin(3x) + ... where a is a real number and |a| < 1. By considering C+iS, show that S = a*sin(x)/(1 - 2a*cos(x) + a^2), and find C.

The infinite series C and S are defined C = acos(x) + a^2cos(2x) + a^3cos(3x) + ..., and S = asin(x) + a^2sin(2x) + a^3sin(3x) + ... where a is a real number and |a| < 1. By considering C+iS, show that S = asin(x)/(1 - 2acos(x) + a^2), and find C.