How do you find the roots, real and imaginary, of #y= 2x^2 + 7x+(2x- 1 )^2 # using the quadratic formula?

Question

1336 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-02T03:50:03

#x=(-3+isqrt15)/12,##(-3-isqrt15)/12#

Solve:

#y=2x^2+7x+(2x-1)^2#

Expand #(2x-1)^2# using the square of a difference: #(a^2-b^2)=a^2-2ab+b^2#, where #a=2x# and #b=1#.

#(2x-1)^2=(2x)^2-2(2x)(1)+1^2#

Simplify.

#(2x-1)^2=4x^2-4x+1#

Add the result to the rest of the equation.

#y=2x^2+7x+4x^2-4x+1#

Simplify.

#y=2x^2+4x^2+7x-4x+1#

#y=6x^2+3x+1#

Substitute #0# for #y#.

#0=6x^2+3x+1# is a quadratic equation in standard form: #a^2+bx+c#, where #a=6#, #b=3#, and #c=1#.

Use the quadratic formula to solve for the roots (values for #x#).

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Substitute the values for #a,b,andc# into the formula.

#x=(-3+-sqrt(3^2-4*6*1))/(2*6)#

Simplify.

#x=(-3+-sqrt(9-24))/12#

#x=(-3+-sqrt(-15))/12#

#x=(-3+-isqrt15)/12#

Roots:

#x=(-3+isqrt15)/12,##(-3-isqrt15)/12#