Question

Solve x2 – 4x – 9 = 29 for x.

Answer 1

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

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=(-b\pm√(b^2-4ac))/(2a)` ...........(1)

Where,
`√(b^2-4ac)` is the discriminant.

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

Substitute in (1) , we get,

`x=(-b\pm√(b^2-4ac))/(2a)`

`\Rightarrow x=(-(-4)\pm√((-4)^2-4\cdot 1 \cdot (-38)))/(2 \cdot 1)`

`\Rightarrow x=(4\pm√(168))/(2)`

`\Rightarrow x_1=(4+√(168))/(2) and \Rightarrow x_2=(4-√(168))/(2)`

Also,
`√(168)=12.96`(approx)

`\Rightarrow x_1=(16.96)/(2) and \Rightarrow x_2=(-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 .

Answer 2

`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 * 1 * -38} }{2 * 1} \\ = (4 \pm √( 16 +152) )/(2) \\ =(4 \pm √( 168) )/(2) \\ =(4 \pm 2√(42) )/(2) \\ =2+√(42) \ or \ 2-√(42)\\=8.48 \ or \ -4.48`