Solve the following quadratic equation(as exact values) using the quadratic formula.
x^2-10x-15=0
I got 5+-2sqrt10/2.
^ Sorry hard to write it out. But basically there is a 5 there, the negative sign is supposed to be under the positive sign, then there is a two outside of the radicand, and inside the radicand is 10, all under 2.
The answer is correct if it was meant to be, after cancelling the two's:
5±√(10)
The answer is correct if it was meant to be (without the 2 in the denominator):
5±2√(10)
Where does that two in the denominator go? Since it has to cancel out with all terms.
x=(-b±sqrt(b²-4ac))/2a
a=1
b=-10
c=-15
so substituting,
x=(10±sqrt((-10)²-4(1)(-15)))/2
=(10±sqrt(160))/2
=(10±4sqrt(10))/2
=5±2sqrt(10)
You may have (like I did) subtracted 4ac (and got 40) instead of adding to get 160.
To solve the quadratic equation x^2 - 10x - 15 = 0, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this equation, a = 1, b = -10, and c = -15.
Substituting these values into the formula, we have:
x = (-(-10) ± sqrt((-10)^2 - 4(1)(-15))) / (2*1)
= (10 ± sqrt(100 + 60)) / 2
= (10 ± sqrt(160)) / 2
To simplify further, we can factor out the square root of 160:
x = (10 ± sqrt(16 * 10)) / 2
= (10 ± 4√10) / 2
Next, we can simplify the expression by dividing both the numerator and the denominator by 2:
x = 5 ± 2√10
Therefore, the exact solutions to the quadratic equation x^2 - 10x - 15 = 0 are x = 5 + 2√10 and x = 5 - 2√10.