MATH SOLVE

5 months ago

Q:
# Solve for x in the equation

Accepted Solution

A:

Answer:The solution is [tex]\displaystyle x=1\pm \sqrt{47}[/tex]. Fourth optionExplanation:Solve for x:[tex]2x^2+3x-7=x^2+5x+39[/tex]Move all the terms from the right to the left side of the equation, a zero in the right side:[tex]2x^2+3x-7-x^2-5x-39=0[/tex]Join all like terms:[tex]x^2-2x-46=0[/tex]The general form of the quadratic equation is:[tex]ax^2+bx+c=0[/tex]Solve the quadratic equation by using the formula:[tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]In our equation: a=1, b=-2, c=-46Substituting into the formula:[tex]\displaystyle x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-46)}}{2(1)}[/tex][tex]\displaystyle x=\frac{2\pm \sqrt{4+184}}{2}[/tex][tex]\displaystyle x=\frac{2\pm \sqrt{188}}{2}[/tex]Since 188=4*47[tex]\displaystyle x=\frac{2\pm \sqrt{4*47}}{2}[/tex]Take the square root of 4:[tex]\displaystyle x=\frac{2\pm 2\sqrt{47}}{2}[/tex]Divide by 2:[tex]\displaystyle x=1\pm \sqrt{47}[/tex]First option: Incorrect. The answer does not matchSecond option: Incorrect. The answer does not matchThird option: Incorrect. The answer does not matchFourth option: Correct. The answer matches exactly this option