How do you solve the quadratic using the quadratic formula given #4x^2+4x-8=1#?

1 Answer
Aug 29, 2016

Answer:

Solution is #(-1-sqrt10)/2# and #(-1+sqrt10)/2#.

Explanation:

For the equation #ax^2+bx+c=0#, the roots are given by #x=(-b+-sqrt(b^2-4ac))/(2a)#. As can be seen type of roots depend upon discriminant #b^2-4ac#. If is negative, roots are complex and if it is a complete square (and coefficients are rational), roots are rational.

In the equation #4x^2+4x-8=1# or #4x^2+4x-9=0#, the discriminant is #(4)^2-4×4×(-9)=16+144=160#, which is not a complete square and hence roots are irrational.

Roots are #x=(-4+-sqrt160)/(2×4)#

= #(-4+-4sqrt10)/8#

= #(-1+-sqrt10)/2#

Hence solution is #(-1-sqrt10)/2# and #(-1+sqrt10)/2#.