Engineering Mechanics - General Principles - Discussion
Discussion Forum : General Principles - General Questions (Q.No. 1)
1.
Solve the following equation for the two roots of x: x2 — 16 = 0
Discussion:
27 comments Page 1 of 3.
Darshan Dodia said:
8 years ago
According to Quadratic Equation ax^2+bx+c=0, roots of equation x = (-b+/-sqrt(b^2-4ac))/(2a).
(1)
Rakesh said:
1 decade ago
(x)2-16 = 0.
We know that a2-b2 = (a+b)(a-b).
(x)2-(4)2 = 0.
(x+4)(x-4) = 0.
x = -4, +4.
We know that a2-b2 = (a+b)(a-b).
(x)2-(4)2 = 0.
(x+4)(x-4) = 0.
x = -4, +4.
Nishita said:
1 decade ago
(x^2-16)=0
(x^2-4^2)=0
(x-4)(x+4)=0
therefore,
(x-4)=0,x=4
(x+4)=0,x=-4
hence,x=4,-4
(x^2-4^2)=0
(x-4)(x+4)=0
therefore,
(x-4)=0,x=4
(x+4)=0,x=-4
hence,x=4,-4
Kamal said:
1 decade ago
Why do use iota. Ioti (i) its value is also -1. That answer is also right option b.
Ajmal k said:
7 years ago
X^2-16=0.
X= +/-(√16),
X= +/-(4),
X=(+4) & x=(-4).
X= +/-(√16),
X= +/-(4),
X=(+4) & x=(-4).
(1)
Prabu said:
1 decade ago
Well Naveen you gave the exact method to find the ans.
Naveen said:
1 decade ago
x^2-16=0
(x-4)*(x+4)=0
(x-4)=0 or (x+4)=0
x=4 , -4
(x-4)*(x+4)=0
(x-4)=0 or (x+4)=0
x=4 , -4
ASHOK said:
1 decade ago
x^2-16=0
x^2=16
x=sq.root 16
i.e.x=+ or - 4
x=4;x=-4
x^2=16
x=sq.root 16
i.e.x=+ or - 4
x=4;x=-4
Lakshmi said:
1 decade ago
x^2 = 16
x = sqrt(16)
Therefore, x = +4 or -4.
x = sqrt(16)
Therefore, x = +4 or -4.
Billa said:
1 decade ago
x^2 - 16 = 0
x^2 = 16
x = sqrt(16)
x = 4 or -4
x^2 = 16
x = sqrt(16)
x = 4 or -4
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers