The equation x3-3x+4=0 has only one real root. What is its first approximate value as obtained by the method of false position in (-3, -2) ?

Question

The equation x3-3x+4=0 has only one real root. What is its first approximate value as obtained by the method of false position in (-3, -2) ? (A)  -2.125  (B)  2.125  (C)  -2.812  (D)  2.812

Tardigrade Admin · Accepted Answer

We have, x2=x0-(x1-x0/f(x1)-f(x0)) Here, x0=-3, x1=-2, f(x)=x3-3x+4 ∴ f(x0)=-27+9+4=-14 f(x1)=-8+6+4=2 So, x2=-3-(1/16)× (-14) =-3+(7/8)=-(17/8)=-2.125

Q. The equation $x^{3} - 3 x + 4 = 0$ has only one real root. What is its first approximate value as obtained by the method of false position in $(- 3, - 2)$ ?

$- 2.125$

$2.125$

$- 2.812$

$2.812$

Q. The equation x3−3x+4=0 has only one real root. What is its first approximate value as obtained by the method of false position in (−3,−2) ?

−2.125

2.125

−2.812

2.812

We have, x2​=x0​−f(x1​)−f(x0​)x1​−x0​​ Here, x0​=−3,x1​=−2,f(x)=x3−3x+4 ∴ f(x0​)=−27+9+4=−14 f(x1​)=−8+6+4=2 So, x2​=−3−161​×(−14) =−3+87​=−817​=−2.125

Q. The equation $x^{3} - 3 x + 4 = 0$ has only one real root. What is its first approximate value as obtained by the method of false position in $(- 3, - 2)$ ?

$- 2.125$

$2.125$

$- 2.812$

$2.812$