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:

Option a is correct

Step-by-step 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 $y=\frac{-1\pm \sqrt{1+12}}{6}$

So,

$x^2=\frac{-1\pm \sqrt{1+12}}{6}=\frac{-1\pm \sqrt{13}}{6}\\x=\pm \sqrt{\left ( \frac{-1\pm \sqrt{13}}{6} \right )}$

Real zeroes:

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