Solve for x in the equation... 2x squared + 3x - 7 = x squared + 5x + 39 a. x = -6 plus-minus the square root of 82 b. x = -6 plus-minus 2 root 17 c. x = 1 plus-minus the root of 33 d. x = 1 plus-minus the root of 47 Please help!! Timed test!!

2 answers
Question:

Solve for x in the equation... 2x squared + 3x - 7 = x squared + 5x + 39
a. x = -6 plus-minus the square root of 82
b. x = -6 plus-minus 2 root 17
c. x = 1 plus-minus the root of 33
d. x = 1 plus-minus the root of 47
Please help!! Timed test!!

Answers

Answer:D) x = 1 ± √(47)Step-by-step explanation:2x² + 3x - 7 = x² + 5x + 39move all terms to one side:x² - 2x - 46 = 0complete the square:x² - 2x + 1 - 1 - 46 = 0(x - 1)² - 47 = 0add 47 to both sides:(x - 1)² = 47square 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 47The 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 47A similar problem is given at https://brainly.com/question/24909950

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