①
√(x² + 3x + 7)

= √(x² + 2(3/2)x + (3/2)² - (3/2)² + 7)

= √(x + 3/2)² + 4¾)


②
√(x² + 3x - 9)

= √(x² + 2(3/2)x + (3/2)² - (3/2)² - 9)

= √((x + 3/2)² - 11¼)


③
Now (x + 3/2)² - 11¼ ≥ 0 in order for √((x + 3/2)² - 11¼) to be a real value.

(Cannot square root a negative value at this level. It has to be 0 or positive.
And this also means √((x + 3/2)² - 11¼) ≥ 0 for all real x)


So (x + 3/2)² - 11¼ ≥ 0
Then (x + 3/2)² ≥ 11¼


④
But if (x + 3/2)² ≥ 11¼,
then (x + 3/2)² + 4¾ ≥ 11¼ + 4¾

(x + 3/2)² + 4¾ ≥ 16
√((x + 3/2)² + 4¾) ≥ √16
√((x + 3/2)² + 4¾) ≥ 4


⑤
Since √((x + 3/2)² + 4¾) ≥ 4 when √((x + 3/2)² - 11¼) ≥ 0,


Then
√((x + 3/2)² + 4¾) + √((x + 3/2)² - 11¼) ≥ 4 for all real x.

Which means that

√(x² + 3x + 7) + √(x² + 3x - 9) ≥ 4 ≠ 2 for all real x.

So it can never be equal 2 and there are no real roots

(edit: there is actually no solution for this equation, not even complex roots)

It is not accurate to say that there will always be an imaginary root to the equation. For this equation there is actually no solution, not even complex/imaginary ones.

What you have done is to equate each quadratic expression to 0 first, and then applied the quadratic formula. But you cannot do this as the equation did not equate either to 0.

√(x² + 3x + 7) is actually real for all real x and exists since it equals :

√((x + 3/2)² + 4¾)

And since (x + 3/2) ≥ 0 for all real x ,
(x + 3/2) + 4¾ ≥ 4¾ > 0

Which means √((x + 3/2)² + 4¾) > 0 for all real x.

Let u = x² + 3x - 9

Then we are trying to solve

sqrt (u + 16) + sqrt u = 2

The difference of 16 between u and u + 16, combined with the fact that u >= 0, makes the equation have a minimum value of 4 for all real values of u. We cannot find any value of x satisfying the equation.

So, the original equation does not have any real roots.

Not accurate to say for all real values of u either. It is strictly for all u ≥ 0 only

Extended : (finding the range of x)


(x + 3/2)² - 11¼ ≥ 0

(x + 3/2)² ≥ 11¼

|x + 3/2| ≥ √11¼

x + 3/2 ≥ √11¼ or x + 3/2 ≤ -√11¼

x ≥ √11¼ - 3/2 or x ≤ -√11¼ - 3/2

x ≥ 3/2 (√5 - 1) or x ≤ -3/2 (√5 + 1)

i'm not sure whether you would use this method, but you could perhaps use the discriminant to show that the result cannot be calculated, as b^2-4ac for both square roots is negative, and negative numbers cannot be squared.

The issue with that is, you are directly equating each quadratic expression to 0, which is not what the equation shows.

Furthermore the discriminant is actually positive for x² + 3x -9 and not negative

i.e
3² - 4(1)(-9) = 45 > 0

How about equating each equation to less than equal to 2?

Since sqrt(x^2+3x+7) + sqrt(x^2+3x-9) =2, either sqrt(x^2+3x+7) or sqrt(x^2+3x-9) must be less than or equal to 2. If one of them is 2 the other will have to be 0.

I’ve tried this and the following are the steps;

sqrt(x^2+3x+7) ≤ 2
x^2+3x+7 ≤ 4
x^2+3x+3 ≤ 0

By b^2-4ac, no real roots (only complex) therefore sqrt(x^2+3x+7) + sqrt(x^2+3x-9) ≠ 2.

The square root cannot be negative
(since we are dealing with principal values eg. √4 = 2 and not -2) so the part where 0 ≤ √(x² + 3x + 7) has to be included as well.


So the inequality should be :

0 ≤ √(x² + 3x + 7) ≤ 2
0 ≤ √((x + 3/2)² + 4¾) ≤ 2

Next,

①
For all real values of x (real roots),
(x + 3/2)² ≥ 0
Then (x + 3/2)² + 4¾ ≥ 4¾ > 4
√((x + 3/2)² + 4¾) ≥ √4¾ > 2

So √(x² + 3x + 7) > 2 for all real roots/real values of x. This in turn means there are no real roots for 0 ≤ √(x² + 3x + 7) ≤ 2


Or

Using discriminant as per your method.

0 ≤ x² + 3x + 7 ≤ 4

Then 0 ≤ x² + 3x + 7 and x² + 3x + 3 ≤ 0

Discriminants are -19 and -3 , which < 0 so there are no real roots.


However it should be noted that even though there are complex roots for each value between 0 and 4 (inclusive) that is equated to x² + 3x + 7 ,

Eg x² + 3x + 7 = 4,
x² + 3x + 7 = 3.5
x² + 3x + 7 = 0.8 etc.

it is not possible to get a range for the values since there are 2 dimensions in the complex number (real and imaginary)

Furthermore, for any complex root satisfying 0 ≤ x² + 3x + 7 ≤ 4, it would result in the following :

-16 ≤ x² + 3x - 9 ≤ -12

So these roots cannot satisfy 0 ≤ x² + 3x - 9 ≤ 4, meaning that there are also no complex roots for the whole equation.


Now for the other one,

0 ≤ √(x² + 3x - 9) ≤ 2 as well since the two square roots have to add up to 2.

0 ≤ ((x + 3/2)² - 11¼) ≤ 4

For all real values of x (real roots),
(x + 3/2)² ≥ 0

Then (x + 3/2)² - 11¼ ≥ -11¼

So there are real values that can satisfy the inequality 0 ≤ x² + 3x - 9 ≤ 4

But this also leads to 16 ≤ x² + 3x + 7 ≤ 20, which will not satisfy 0 ≤ x² + 3x + 7 ≤ 4.

So overall, there are no real or complex roots satisfying the whole equation.