Verified answer

a)

x² + 10x + 36 = (x² + 10x + 25) + 11 = (x + 5)² + 11

(x + 5)² + 11 = 0

b)

(x + 5)² = - 11 is impossible in ℝ

no real number when is squared gives negative result

(complex numbers like 5i when squared give negative result

(5i)² = -25

because by definition i² = -1

"real", "complex" or "rational" are technical words which have nothing to do with their meaning in non-mathematical context

a)

another way

is based on the identity principle of polynomials:

two polynomials are identical if:

- they have the same degree;

- terms of corresponding degree have the same coefficient

x² + 10x + 36 = (x + a)² + b

expand right hand side

x² + 10x + 36 = x² + 2ax + a² + b

to be identical must be

2a = 10

a² + b = 36

a = 5

25 + b = 36

b = 11

c)

quadratic equations

ax² + bx + c = 0

define the discriminant

Δ = b² - 4ac

roots are

x₁ = (-b - √Δ)/(2a)

x₂ = (-b + √Δ)/(2a)

they can be:

two distinct roots if Δ > 0

two equal roots if Δ = 0

no real root if Δ < 0

if the equation x² + 10x + k = 0 has equal roots

this means that Δ = 0,

Δ = 10² - 4k = 100 - 4k

100 - 4k = 0 if

- 4k = -100

k = 25

if you replace k = 25 in the equation, you get

x² + 10x + 25 = 0

Δ = 10² - 4⋅25 = 0

roots are

x₁ = (-10 - √0)/2 = -5

x₂ = (-10 + √0)/2 = -5

x₁ = x₂ = -5

I hope this is useful to you

Roots of a function are the values of x where the function touches or crosses the x-axis. For example, y = x^2 touches the x-axis at x = 0, but y = x^2 + 2 never does, and so the roots of the equation are imaginary, not real.

One easy way to find if a function has real roots is to use the determinant. This is the part of the quadratic equation that is under the radical: b^2 - 4ac.

Here, b^2 - 4ac = 10^2 - 4 * 1 * 36 = 100 - 144 = -44.

You know that you can't take the square root of a negative number, or that if you do, you get imaginary answers. sqrt(4) = 2, and sqrt(-4) = 2i.

Because solving the quadratic equation in this case will force us to take the square root of -44, any answer is going to be imaginary.

We can also find this out by completing the square.

10/2 = 5

5^2 = 25

(x^2 + 10) + 36

(x^2 + 10x + 25) + 36 - 25

(x + 5)^2 + 11

Now remember that this equation is being solved by being set equal to 0.

So (x + 5)^2 + 11 = 0

(x + 5)^2 = -11

x + 5 = sqrt(-11)

And you can see that we're back to the same problem. In fact, if we took the square root of -44, we'd get 2 * sqrt(11), and since the 2a in the denominator would be 2 * 1 = 2, the 2s would cancel out, and we'd get sqrt(-11).

***

A quadratic equation is said to have equal roots when the two roots are the same. On a graph, it means the function touches, but does not cross, the x-axis. In factors, it means the function factors to (x +/- a)^2

Completing the square as before we get (x + 5)^2 + k - 25 = 0

So when k = 25, the function will be (x + 5)^2 = 0, and both roots will be x = -5. And the graph meets the x-axis at (-5,0).

To find the point where this graph meets the y axis, plug in 0 for x

0^2 + 10 * 0 + 25 = 25, so the graph intersects the y-axis at (0,25)