Answer:
D) x = 1 ± √(47)
Step-by-step explanation:
2x² + 3x - 7 = x² + 5x + 39
move all terms to one side:
x² - 2x - 46 = 0
complete the square:
x² - 2x + 1 - 1 - 46 = 0
(x - 1)² - 47 = 0
add 47 to both sides:
(x - 1)² = 47
square root both sides:
x - 1 = ±√(47)
add 1 to both sides:
x = 1 ± √(47)
The solution of the quadratic equation is:
d. x = 1 plus-minus the root of 47
The equation given is:
[tex]2x^2 + 3x - 7 = x^2 + 5x + 39[/tex]
Passing everything to the same side, we have that:
[tex]2x^2 - x^2 + 3x - 5x - 7 - 39 = 0[/tex]
[tex]x^2 - 2x - 46 = 0[/tex]
Which is a quadratic equation with coefficients [tex]a = 1, b = -2, c = -46[/tex]. Then:
[tex]\Delta = b^2 - 4ac = (-2)^2 - 4(1)(-46) = 188[/tex]
[tex]x_{1} = \frac{(-2) + \sqrt{188}}{2} = \frac{2 + 2\sqrt{47}}{2} = 1 + \sqrt{47}[/tex]
[tex]x_{2} = \frac{-(-2) - \sqrt{188}}{2} = \frac{2 - 2\sqrt{47}}{2} = 1 - \sqrt{47}[/tex]
Thus, the solution is:
d. x = 1 plus-minus the root of 47
A similar problem is given at https://brainly.com/question/24909950
