Question

Solve x2 – 4x – 9 = 29 for x.

Answer 1

$x^2 - 4x - 9 = 29 \\ x^2 - 4x - 9 - 29 = 0 \\ x^2 - 4x - 38 = 0 \\ x= \frac{-b \pm \sqrt{ b^{2} -4ac} }{2a}$; where a = 1, b = -4 and c = -38
$x= \frac{-(-4) \pm \sqrt{ (-4)^{2} -4 \times 1 \times -38} }{2 \times 1} \\ = \frac{4 \pm \sqrt{ 16 +152} }{2} \\ =\frac{4 \pm \sqrt{ 168} }{2} \\ =\frac{4 \pm 2\sqrt{42} }{2} \\ =2+\sqrt{42} \ or \ 2-\sqrt{42}\\=8.48 \ or \ -4.48$

Answer 2

Answer:

The two root of the given quadratic equation  $x^2-4x-9=29$ is 8.48 and -4.48 .

Step-by-step explanation:

Consider, the given Quadratic equation, $x^2-4x-9=29$

Thus on simplification, we get,

$x^2-4x-9-29=0 \Rightarrow x^2-4x-38=0$

We can solve using quadratic formula,

For a given quadratic equation $ax^2+bx+c=0$ we can find roots using,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$  ...........(1)

Where, $\sqrt{b^2-4ac}$ is the discriminant.

Here, a = 1 , b = -4 , c = -38

Substitute in (1) , we get,

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

$\Rightarrow x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot 1 \cdot (-38)}}{2 \cdot 1}$

$\Rightarrow x=\frac{4\pm\sqrt{168}}{2}$

$\Rightarrow x_1=\frac{4+\sqrt{168}}{2} and \Rightarrow x_2=\frac{4-\sqrt{168}}{2}$

Also, $\sqrt{168}=12.96$(approx)

$\Rightarrow x_1=\frac{16.96}{2} and \Rightarrow x_2=\frac{-8.96}{2}$

$\Rightarrow x_1=8.48 and \Rightarrow x_2=-4.48$

Thus, the two root of the given quadratic equation  $x^2-4x-9=29$ is 8.48 and -4.48 .