Question

Solve for x in the equation x squared minus 4 x minus 9 = 29. x = 2 plus-or-minus StartRoot 42 EndRoot x = 2 plus-or-minus StartRoot 33 EndRoot x = 2 plus-or-minus StartRoot 34 EndRoot x = 4 plus-or-minus StartRoot 42 EndRoot

Answer 1

Answer is A or the first one

Answer 2

`x=2$\pm$√(42)`

Explanation:

The given equation is:

`x^(2) -4x-9=29\\\Rightarrow x^(2) -4x-9-29=0\\\Rightarrow x^(2) -4x-38=0`

Formula:

A quadratic equation
`ax^(2) +bx+c=0` has the following roots:

`x=(-b+\sqrt D)/(2a)\ and\\x=(-b-\sqrt D)/(2a)`

Where
`D= b^(2) -4ac`

Comparing the equation with
`ax^(2) +bx+c=0`

a = 1

b = -4

c= -38

Calculating D,

`D= (-4)^(2) -4(1)(-38)\\\Rightarrow D = 16+152 = 168`

Now, finding the roots:

`x=\frac{-(-4)+\sqrt {168}}{2* 1}\\\Rightarrow x=\frac{4+2\sqrt {42}}{2}\\\Rightarrow x=2+\sqrt {42}\\and\\x=\frac{-(-4)-\sqrt {168}}{2* 1}\\\Rightarrow x=\frac{4-2\sqrt {42}}{2}\\\Rightarrow x=2-\sqrt {42}`

So, the solution is:

`x=2$\pm$√(42)`