The degree is two so it must have more than one solution.
The given equation is
Let us use the square root method to solve this equation.
Since the degree of x is two, the fundamental theorem of algebra says it must have 2 roots.
We take square root to obtain,
Split the plus or minus sign to obtain,
48/40 = 1 8/40
1 8/40 Simplify to 1 1/5
Hope this helps!
Answers:k = 13The smallest zero or root is x = -10
note: you can write "x^2" to mean "x squared"
f(x) = x^2+3x-10
f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5
f(x+5) = (x^2+10x+25)+3(x+5)-10
f(x+5) = x^2+10x+25+3x+15-10
f(x+5) = x^2+13x+30
Compare this with x^2+kx+30 and we see that k = 13
Factor and solve the equation below
x^2+13x+30 = 0
(x+10)(x+3) = 0
x+10 = 0 or x+3 = 0
x = -10 or x = -3
The smallest zero is x = -10 as its the left-most value on a number line.