If f(x)=(x^2+3x)(x^2+3x+2) , then the sum of all real roots of the equation \\
f(x)+1
$ = 9701 , is⌨️ Shortcuts:[1-4 / A-D] Select[Enter] Check[B] Star[S] Solution
Official Correct Answer: C. The question says: If \( f(x) = (x^2 + 3x)(x^2 + 3x + 2) \), then the sum of all real roots of the equation \( \sqrt{f(x) + 1} = 9701 \) is... and then there are four options: A) -6, B) 6, C) -3, D) 3.
Okay, so first, We find the real roots of the equation \( \sqrt{f(x) + 1} = 9701 \), and then sum them up. Let's break this down.
Starting with the equation \( \sqrt{f(x) + 1} = 9701 \). To get rid of the square root, I can square both sides. That should make things easier to handle. So, squaring both sides gives:
\( f(x) + 1 = (9701)^2 \)
Hmm, that's a big number. Compute \( 9701^2 \). Wait, maybe I don't need to compute it exactly right now. if I can express \( f(x) \) in a simpler form first.
Looking at \( f(x) = (x^2 + 3x)(x^2 + 3x + 2) \). Hmm, that looks like a product of two quadratics. Maybe I can let \( y = x^2 + 3x \) to simplify things. Let me try that substitution.
Let \( y = x^2 + 3x \). Then, \( f(x) = y(y + 2) = y^2 + 2y \).
So, substituting back into the equation \( f(x) + 1 = 9701^2 \), we get:
\( y^2 + 2y + 1 = 9701^2 \)
Wait, \( y^2 + 2y + 1 \) is a perfect square. That's \( (y + 1)^2 \). So, the equation becomes:
\( (y + 1)^2 = 9701^2 \)
Taking square roots on both sides, we have:
\( y + 1 = \pm 9701 \)
So, two cases:
1. \( y + 1 = 9701 \) => \( y = 9700 \)
2. \( y + 1 = -9701 \) => \( y = -9702 \)
Now, remember that \( y = x^2 + 3x \). So, substituting back, we have two quadratic equations:
1. \( x^2 + 3x = 9700 \)
2. \( x^2 + 3x = -9702 \)
Let me write them down:
1. \( x^2 + 3x - 9700 = 0 \)
2. \( x^2 + 3x + 9702 = 0 \)
Now, We find the real roots of these quadratics and then sum them up.
Starting with the first equation: \( x^2 + 3x - 9700 = 0 \)
Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where a=1, b=3, c=-9700.
Calculating the discriminant:
\( D = 3^2 - 4(1)(-9700) = 9 + 38800 = 38809 \)
Hmm, what's the square root of 38809? . 197^2 is 38809 because 200^2 is 40000, so 197^2 is 38809. So, \sqrt{38809} = 197.
Therefore, the roots are:
\( x = \frac{-3 \pm 197}{2} \)
Calculating both roots:
1. \( x = \frac{-3 + 197}{2} = \frac{194}{2} = 97 \)
2. \( x = \frac{-3 - 197}{2} = \frac{-200}{2} = -100 \)
So, the roots from the first equation are 97 and -100.
Now, moving on to the second equation: \( x^2 + 3x + 9702 = 0 \)
Again, using the quadratic formula with a=1, b=3, c=9702.
Discriminant:
\( D = 3^2 - 4(1)(9702) = 9 - 38808 = -38800 - 1 = -38800 -1? Wait, 9 - 38808 is -38800 - 1? Wait, 9 - 38808 is 9 - 38808 = -38799.
Wait, that's negative. So, discriminant is negative, which means no real roots for this equation. So, only the first equation contributes real roots.
Therefore, the real roots are 97 and -100.
Now, the question asks for the sum of all real roots. So, 97 + (-100) = -3.
Looking at the options, C is -3. So, the answer should be C.
1. Started with \( \sqrt{f(x) + 1} = 9701 \), squared both sides to get \( f(x) + 1 = 9701^2 \).
2. Expressed \( f(x) \) as \( y(y + 2) \) where \( y = x^2 + 3x \), leading to \( (y + 1)^2 = 9701^2 \).
3. Solved for y, getting y = 9700 and y = -9702.
4. Substituted back to get two quadratics, solved the first one, found roots 97 and -100.
5. Second quadratic had negative discriminant, so no real roots.
6. Sum of real roots: 97 + (-100) = -3.
Everything seems to check out. So, yeah, the answer is C) -3.
**Final Answer**
The sum of all