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erastova [34]
3 years ago
11

Solve for x in the equation.

Mathematics
1 answer:
labwork [276]3 years ago
8 0

Answer:

The solution is \displaystyle x=1\pm \sqrt{47}. Fourth option

Explanation:

Solve for x:

2x^2+3x-7=x^2+5x+39

Move all the terms from the right to the left side of the equation, a zero in the right side:

2x^2+3x-7-x^2-5x-39=0

Join all like terms:

x^2-2x-46=0

The general form of the quadratic equation is:

ax^2+bx+c=0

Solve the quadratic equation by using the formula:

\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

In our equation: a=1, b=-2, c=-46

Substituting into the formula:

\displaystyle x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-46)}}{2(1)}

\displaystyle x=\frac{2\pm \sqrt{4+184}}{2}

\displaystyle x=\frac{2\pm \sqrt{188}}{2}

Since 188=4*47

\displaystyle x=\frac{2\pm \sqrt{4*47}}{2}

Take the square root of 4:

\displaystyle x=\frac{2\pm 2\sqrt{47}}{2}

Divide by 2:

\displaystyle x=1\pm \sqrt{47}

First option: Incorrect. The answer does not match

Second option: Incorrect. The answer does not match

Third option: Incorrect. The answer does not match

Fourth option: Correct. The answer matches exactly this option

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