Solve the equation 2x^2+15x+38=x^2 to the nearest tenth.
x²+15x+38=0
x=(-15±√(225-152)/2
x=(-15±√73)/2
2x^2+15x+38=x^2
subtracting x^2 ... x^2 + 15x + 38 = 0
use the quadratic formula to find the two solutions
To solve the equation, we first combine like terms to get:
2x^2 + 15x + 38 = x^2
Next, we subtract x^2 from both sides of the equation to have all the terms on one side:
2x^2 - x^2 + 15x + 38 = 0
Simplifying this, we get:
x^2 + 15x + 38 = 0
To find the value of x, we can use the quadratic formula, which is:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = 15, and c = 38. Substituting these values into the quadratic formula, we have:
x = (-(15) ± √((15)^2 - 4(1)(38))) / (2(1))
Simplifying this further, we get:
x = (-15 ± √(225 - 152)) / 2
x = (-15 ± √(73)) / 2
Therefore, the solutions to the equation are approximately:
x = (-15 + √(73)) / 2 ≈ -4.536
x = (-15 - √(73)) / 2 ≈ -10.464
Thus, the approximate solutions to the equation 2x^2 + 15x + 38 = x^2 to the nearest tenth are x ≈ -4.5 and x ≈ -10.5.
To solve the equation, we want to isolate the variable x on one side of the equation. Let's rearrange the equation by moving all the terms to one side:
2x^2 + 15x + 38 = x^2
Subtract x^2 from both sides:
2x^2 - x^2 + 15x + 38 = 0
Simplifying the equation gives:
x^2 + 15x + 38 = 0
Now, we can either use the quadratic formula or try factoring the equation to find the solutions.
Let's try factoring the equation first. We need to find two numbers that multiply to give 38 and add up to 15.
The numbers that satisfy this condition are 2 and 19. Rewriting the middle term using these numbers, we have:
x^2 + 2x + 19x + 38 = 0
Now, we can factor by grouping:
(x^2 + 2x) + (19x + 38) = 0
Taking out the common factor, we get:
x(x + 2) + 19(x + 2) = 0
Now, we have a common binomial factor of (x + 2):
(x + 2)(x + 19) = 0
To find the solutions, we set each factor equal to zero:
x + 2 = 0 or x + 19 = 0
Solving for x gives:
x = -2 or x = -19
Therefore, the solutions to the equation are x = -2 or x = -19.
To find the solutions to the nearest tenth, we do not change the values of -2 and -19 as they are already in decimal form rounded to the nearest tenth, if necessary.