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:

$x=2 \pm \sqrt{42}$

Step-by-step 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:

$x=\dfrac{-b+\sqrt D}{2a}\ and\\x=\dfrac{-b-\sqrt D}{2a}$

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:

$x=\dfrac{-(-4)+\sqrt {168}}{2\times 1}\\\Rightarrow x=\dfrac{4+2\sqrt {42}}{2}\\\Rightarrow x=2+\sqrt {42}\\and\\x=\dfrac{-(-4)-\sqrt {168}}{2\times 1}\\\Rightarrow x=\dfrac{4-2\sqrt {42}}{2}\\\Rightarrow x=2-\sqrt {42}$

So, the solution is:

$x=2 \pm \sqrt{42}$