Question

Find the discriminant.

4w 2 + 8W + 4 = 0
How many real solutions does the equation have?
no real
solutions
one real
solution
two real
solutions

Answer 1

one real solution being only -1

Explanation:

Solve for w over the real numbers:

4 w^2 + 8 w + 4 = 0

Divide both sides by 4:

w^2 + 2 w + 1 = 0

Write the left hand side as a square:

(w + 1)^2 = 0

Take the square root of both sides:

w + 1 = 0

Subtract 1 from both sides:

Answer: |

| w = -1

Answer 2

`Discriminant=0\\b)\ One\ Real\ Solution`

Explanation:

`We\ are\ given\ that:\\4w^2+8w+4=0\\We\ know\ every\ quadratic\ equation\ can\ be\ represented\ in\ the\ form\ of:\\ax^2+bx+c=0,\ where\ x \\eq 0\\Hence,\\Lets\ first\ recognize\ the\ co-efficients\ of\ the\ algebraic\ terms( x^2,\ x,\ x^0),\\ which\ are\ a,b\ and\ c\ respectively.`

`Hence,\\4w^2+8w+4=0\\4(w^2)+8(w)+4(w^0)=0\\Hence,\\a=4,\ b=8,\ and\ c=4.`

`Now,\\We\ can\ decide\ the\ nature\ of\ roots\ the\ quadratic\ equation\ has,\ by\ looking\\ at\ its\ Discriminant.\\Hence,\\Lets\ first\ find\ the\ Discriminant\ for\ our\ equation:\\`

`We\ know\ that,\\Discriminant=b^2-4ac\\\\Substituting\ a=4, b=8, c=4\ in\ the\ Discriminant\ Formula,\ we\ get:\\8^2-4*4*4\\=64-64\\=0\\Hence,\\Discriminant=0`

`We\ also\ know\ that\ if,\\D>0, The\ equation\ forms\ 2\ distinct\ real\ roots.\\D=0,The\ equation\ forms\ only\ one\ real\ root\ exactly\ at\ the\ x-axis.\\D<0,\ The\ equation\ forms\ Imaginary/Complex\ Roots\ or\ basically\ no\ real\ roots.`
`Here,\\As\ D=0,\\The\ equation\ forms\ only\ one\ Real\ root\ or\ has\ only\ one\ solution.`