Question

Approximate the real zeros f(x)=3x^4+x^2-1 to the nearest tenth

a. -0.7,0.7
b.-0.5,0.5
c.-7,7
d.-0.6,0.6

Answer 1

Answer:a. -0.7, 0.7

Explanation:

Answer 2

Option a is correct

Explanation:

Given:
`3x^4+x^2-1`

To find: roots of the equation

Solution:

A number x is a root of an equation if it satisfies the equation. It is a real root if it is also a real number.

`3x^4+x^2-1`

Take
`y=x^2`

`3x^4+x^2-1=3y^2+y-1`

For an equation of the form
`ay^2+by+c=0`, roots are given by
`y=(-1\pm √(1+12))/(6)`

So,

`x^2=(-1\pm √(1+12))/(6)=(-1\pm √(13))/(6)\\x=\pm \sqrt{\left ( (-1\pm √(13))/(6) \right )}`

Real zeroes:

`x=\pm \sqrt{\left ( (-1+√(13))/(6) \right )}\\=\pm \sqrt{\left ( (-1+ 3.61)/(6) \right )}\\=\pm \sqrt{\left ( (2.61)/(6) \right )} \\=\pm √(0.44)\\=\pm 0.7`