Approximate the real zeros of f(x)= 3x^4 + x^2 - 1 to the nearest tenth. a. -0.7 , 0.7 c. -7, 7 b. -0.5, 0.5 d. -0.6, 0.6 Please select the best answer from the choices provided
A
B
C
D

x ≈ - 0.7 ≈ 0.7; Option A

Step-by-step explanation:

Let us substitute f ( x ) with 0, as to find the " real zero value( s ) " of the function;

0 = 3x^4 + x^2 - 1 ⇒ Swap sides, as such,

3x^4 + x^2 - 1 = 0 ⇒ Rewrite equation with u being = x^2, u^2 = x^2,

3u^2 + u - 1 = 0 ⇒ And solve for u by completing the square,

3u^2 + u = 1 ⇒ Divide either side by 3,

u^2 + u / 3 = 1 / 3 ⇒ Write equation in the form x^2 + 2ax + a^2 = ( x + a )^2, solving for a,

2au = 1 / 3u,

a = ( 1 / 3u ) / 2u,

a = 1 / 6 ⇒ Add value of a^2 to either side of equation,

u^2 + u / 3 + ( 1 / 6 )^2 = 13 / 36,

( u + 1/ 6 )^2 = 13 / 36 ⇒ Solve for u,

u = ( √13 - 1 ) / 6, and u = ( - √13 - 1 ) / 6 ⇒ Substitute value of u back into first equation to solve for x,

x = √( - 1 + √13 ) / 6 ), x = - √( - 1 + √13 ) / 6 ), x = √( - 1 - √13 ) / 6 ), x = - √( - 1 - √13 ) / 6 ) ⇒ Convert to approximate decimal form,

x = 0.6589 = - 0.6589

Answer; x ≈ - 0.7 ≈ 0.7; Option A

here

step-by-step explanation: