If 2 roots of an equation x^2-3x-3=0 are m and n, what is the value of m^2+n^2?
A. 10
B. 12
C. 15
D. 24
E. 25
The OA is C.
In this PS question I just need to find the roots and then solve the equation that is function of m and n, right? Can I find it using the second grade formula?
Experts, can you give me any sugestion about how can I solve this PS question please? Thanks.
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If 2 roots of an equation x^2-3x-3=0 are m and n, what is...
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Given x² + bx + c = 0:
-b = the SUM of the two roots.
c = the PRODUCT of the two roots
Thus:
m+n = -b = -(-3) = 3.
mn = c = -3.
(m+n)² = m² + n² + 2mn.
Substituting m+n = 3 and mn = -3 into (m+n)² = m² + n² + 2mn, we get:
3² = m² + n² + 2(-3)
9 = m² + n² - 6
15 = m² + n².
The correct answer is C.
-b = the SUM of the two roots.
c = the PRODUCT of the two roots
In the equation above, b = -3 and c = -3.AAPL wrote:If 2 roots of an equation x^2-3x-3=0 are m and n, what is the value of m^2+n^2?
A. 10
B. 12
C. 15
D. 24
E. 25
Thus:
m+n = -b = -(-3) = 3.
mn = c = -3.
(m+n)² = m² + n² + 2mn.
Substituting m+n = 3 and mn = -3 into (m+n)² = m² + n² + 2mn, we get:
3² = m² + n² + 2(-3)
9 = m² + n² - 6
15 = m² + n².
The correct answer is C.
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Hi AAPL,If 2 roots of an equation x^2-3x-3=0 are m and n, what is the value of m^2+n^2?
A. 10
B. 12
C. 15
D. 24
E. 25
The OA is C.
Let's take a look at your question.
$$x^2-3x-3=0$$
For a quadratic equation, we know that,
Sum of the roots = -b= -(-3) = 3
Product of the roots = c = -3
If roots of the given quadratic equation are m and n, then:
$$m+n=3$$
$$mn=-3$$
Using the polynomial identity:
$$\left(m+n\right)^2=m^2+n^2+2mn$$
Plugin in the known values in above equation.
$$\left(3\right)^2=m^2+n^2+2\left(-3\right)$$
$$9=m^2+n^2-6$$
$$m^2+n^2=9+6$$
$$m^2+n^2=15$$
Therefore, Option C is correct.
Hope it helps.
I am available if you'd like any follow up.
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Hi AAPL,
We're told that the 2 roots of the equation x^2-3x-3=0 are M and N. We're asked for the value of M^2 + N^2. If you don't immediately recognize the Algebraic patterns involved, you can still use a bit of 'brute force' arithmetic (and the 'spread' of the answer choices) to get to the correct answer.
To start, we know that there are 2 solutions - and the equation is relatively 'simple', so the solutions are likely 'close' to single-digit integers. We can do a few calculations to look for a pattern:
IF...
X = -2... X^2 - 3X - 3 = 7
X = -1... = 1
X = 0... = -3
X = 1... = -5
X = 2... = -5
X = 3... = -3
X = 4... = 1
Looking at this pattern, the 2 solutions will be (between -1 and 0, but closer to -1) and (between 3 and 4, but closer to 4). (-1)^2 + (4)^2 = 17. There's only one answer that's close to that....
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that the 2 roots of the equation x^2-3x-3=0 are M and N. We're asked for the value of M^2 + N^2. If you don't immediately recognize the Algebraic patterns involved, you can still use a bit of 'brute force' arithmetic (and the 'spread' of the answer choices) to get to the correct answer.
To start, we know that there are 2 solutions - and the equation is relatively 'simple', so the solutions are likely 'close' to single-digit integers. We can do a few calculations to look for a pattern:
IF...
X = -2... X^2 - 3X - 3 = 7
X = -1... = 1
X = 0... = -3
X = 1... = -5
X = 2... = -5
X = 3... = -3
X = 4... = 1
Looking at this pattern, the 2 solutions will be (between -1 and 0, but closer to -1) and (between 3 and 4, but closer to 4). (-1)^2 + (4)^2 = 17. There's only one answer that's close to that....
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
