Let's first write the equation given
x²+6x-13=0
They have told us to use "Completing the square" method.
So we need write x²+6x-13 as a whole square.
x²+6x-13=0
x²+(2*3*x)-9-4 = 0
x²+(2*3*x)-3²=4
Now we can write the LHS as:
(x - 3)²=2²
Hence,
x-3 can either be 2 or -2
Now,
x - 3=2
x=5
Or,
x - 3=-2
x=1
Therefore now we know that x can either be 1 or 5. So the correct answer is Option B.
Answer:
C
Step-by-step explanation:
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The correct answer is C.
You can tell this by factoring the equation to get the zeros. To start, pull out the greatest common factor.
f(x) = x^4 + x^3 - 2x^2
Since each term has at least x^2, we can factor it out.
f(x) = x^2(x^2 + x - 2)
Now we can factor the inside by looking for factors of the constant, which is 2, that add up to the coefficient of x. 2 and -1 both add up to 1 and multiply to -2. So, we place these two numbers in parenthesis with an x.
f(x) = x^2(x + 2)(x - 1)
Now we can also separate the x^2 into 2 x's.
f(x) = (x)(x)(x + 2)(x - 1)
To find the zeros, we need to set them all equal to 0
x = 0
x = 0
x + 2 = 0
x = -2
x - 1 = 0
x = 1
Since there are two 0's, we know the graph just touches there. Since there are 1 of the other two numbers, we know that it crosses there.
The value of the digit 5 in this number is 50 because 5 is in the tenth place.
