Find the roots of the given equation by completing the square.
αx^2 + βx + δ = 0
Do you mean ax^2+ bx+c=0 ??
I think that's safe. The symbols used don't really matter. It's easier to type Roman letters, so I'll use those
ax^2 + bx + c = 0
a(x^2 + b/a x) = -c
a(x^2 + b/a x + (b/2a)^2) = (b/2a)^2 - c
a(x + (b/2a))^2 = (b^2 - 4ac)/4a
(x + b/2a)^2 = (b^2 - 4ac)/4a^2
x + b/2a = ±√(b^2 - 4ac)/2a
x = [-b ± √(b^2 - 4ac)]/2a
Look familiar?
post it.
To find the roots of the given equation by completing the square, follow these steps:
Step 1: Make sure the coefficient of the x^2 term, α, is equal to 1. If it is not, divide the entire equation by α.
Step 2: Move the constant term, δ, to the other side of the equation to form a perfect square trinomial on the left side.
αx^2 + βx = -δ
Step 3: Take half of the coefficient of the x term, β/2, and square it. Add this value to both sides of the equation.
αx^2 + βx + (β/2)^2 = -δ + (β/2)^2
Simplifying, we get:
αx^2 + βx + (β^2/4α) = -δ + (β^2/4α)
Step 4: Rewrite the left side of the equation as a perfect square trinomial.
(√αx + β/(2√α))^2 = (β^2 - 4αδ)/(4α)
Step 5: Take the square root of both sides of the equation.
√(√αx + β/(2√α))^2 = ±√((β^2 - 4αδ)/(4α))
Simplifying, we get:
√αx + β/(2√α) = ±√((β^2 - 4αδ)/(4α))
Step 6: Isolate x by subtracting β/(2√α) from both sides of the equation.
√αx = -β/(2√α) ± √((β^2 - 4αδ)/(4α))
Step 7: Divide both sides of the equation by √α.
x = (-β ± √(β^2 - 4αδ))/(2α)
Now, the equation is in the form x = (-b ± √(b^2 - 4ac))/(2a), which represents the roots of the quadratic equation αx^2 + βx + δ = 0. The roots of the equation are the values of x that satisfy the equation.