Using factorization method;
x²+13x+40=0
x²+13+40
Find two numbers that you can add together to give the co-efficient of variable x. (I.e 13 for this question). Also, you'll find two numbers that you can multiply with each other to give you the whole number as an answer. (I.e 40 in this question). The two numbers must be the same (i.e the two numbers that will be added to give the co-efficient of x and the two numbers that will be multiplied to give the whole number must be the same two).
The two numbers are +5 and +8
The equation will therefore be = x²+5x+8x+40
You'll then factorize (I.e use a common factor of both values to bracket them)
x(x+5)+8(x+5)
(x+8)(x+5)
x is therefore (x+8=0) or (x+5=0)
x=(x=0-8) or (x=0-5)
x= -8 or x= -5
15 & 27 would be the correct pair
<h3>
Answer: 1</h3>
where x is nonzero
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Explanation:
We'll use two rules here
- (a^b)^c = a^(b*c) ... multiply exponents
- a^b*a^c = a^(b+c) ... add exponents
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The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.
Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)
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After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2), (b^2-c^2) and (c^2-a^2)
Add up those exponents (using rule 2 above) and we get
(a^2-b^2)+(b^2-c^2)+(c^2-a^2)
a^2-b^2+b^2-c^2+c^2-a^2
(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)
0a^2 + 0b^2 + 0c^2
0+0+0
0
All three exponents add to 0. As long as x is nonzero, then x^0 = 1
Addition is the first step in solving the equation. Here is how to solve this problem:
x/3 - 3 = 11
+ 3 + 3
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x
(3) -------- = 14(3)
3
x = 42
