Hello all - I would like to check my work on 2 problems.
Solve by factoring :
x^2-13x= -12
x^2-13x+12= -12 + 12
x^2-13x+12= 0
(x-1)(x-12)= 0
x = 1 , 12
Solve using the square root property:
2x^2-35=15
2x^2-35-15=0
2x^2-50=0
2(x-5)(x+5)=0
x = 5 , -5 Not sure I did this one right.
Both sets of answers are correct. I only wonder if in the second one, you should have used the sq. root property. If you did it that way you would get the exact same answer. I just don't want you to get marked down for not following directions.
here's how to do it using sq. roots:
2x^2-35=15
2x^2=50
x^2=25 <-- sq. root both sides
x=+/- 5
Thankyou. I will use the square root way.
Wierd how I got the same answer.
You're welcome.
If you think about it, it makes since why you got the same answer. The way you did it was right, just not the method they were asking for. There is never just one way to do things in math. :-)
To check your work on the first problem, let's simplify the given equation step by step:
x^2 - 13x = -12
First, move all terms to one side of the equation to set it equal to zero:
x^2 - 13x + 12 = 0
Next, factorize the quadratic equation:
(x - 1)(x - 12) = 0
Now, set each factor equal to zero:
x - 1 = 0 or x - 12 = 0
Solve for x in each equation:
x = 1 or x = 12
Therefore, your solutions are correct: x = 1 and x = 12.
For the second problem, let's verify your work:
2x^2 - 35 = 15
Move all terms to one side:
2x^2 - 50 = 0
Now, factorize the quadratic equation:
2(x - 5)(x + 5) = 0
To solve for x, set each factor equal to zero:
x - 5 = 0 or x + 5 = 0
Solve for x in each equation:
x = 5 or x = -5
You have already correctly found the solutions: x = 5 and x = -5.
In summary, the solutions you found for both problems are indeed correct. Keep in mind that factoring and solving quadratic equations using the square root property are valid methods to find the solutions.