Question

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

x^2 + 4 = 6x

Answer 1

The final answers are x = 5.24 OR x = 0.76

Explanation:

Given the equation is x^2 +4 = 6x

Rewriting it in quadratic form as:- x^2 -6x 4 = 0.

a = 1, b = -6, c = 4.

Using Quadratic formula as follows:- x = ( -b ± √(b² -4ac) ) / (2a)

x = ( 6 ± √(36 -4*1*4) ) / (2*1)

x = ( 6 ± √(36 -16) ) / (2)

x = ( 6 ± √(20) ) / (2)

x = ( 6 ± 4.47 ) / (2)

x = (6+4.47) / (2) OR x = (6-4.47) / (2)

x = 10.47/2 OR x = 1.53/2

x = 5.235 OR x = 0.765

Hence, final answers are x = 5.24 OR x = 0.76

Answer 2

Thus, the two root of the given quadratic equation
`x^2+4=6x` is 5.24 and 0.76 .

Explanation:

Consider, the given Quadratic equation,
`x^2+4=6x`

This can be written as ,
`x^2-6x+4=0`

We have to 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 = -6 , c = 4

Substitute in (1) , we get,

`x=(-b\pm√(b^2-4ac))/(2a)`

`\Rightarrow x=(-(-6)\pm√((-6)^2-4\cdot 1 \cdot (4)))/(2 \cdot 1)`

`\Rightarrow x=(6\pm√(20))/(2)`

`\Rightarrow x=(6\pm 2√(5))/(2)`

`\Rightarrow x={3\pm √(5)}`

`\Rightarrow x_1={3+√(5)}` and
`\Rightarrow x_2={3-√(5)}`

We know
`√(5)=2.23607`(approx)

Substitute, we get,

`\Rightarrow x_1={3+2.23607}`(approx) and
`\Rightarrow x_2={3-2.23607}`(approx)

`\Rightarrow x_1={5.23607}`(approx) and
`\Rightarrow x_2=0.76393}`(approx)

Thus, the two root of the given quadratic equation
`x^2+4=6x` is 5.24 and 0.76 .