Without solving determine the number of real solutions for each quadratic equation
1 answer:
Step-by-step explanation:
b^2-4b+3=0
b²-3x-b+3=0
b(b-3)-1(b-3)=0
(b-3)(b-1)=0
either
b=3 or b=1
.
2n^2 + 7 = -4n + 5
2n²+4n+7-5=0
2n²+4n+2=0
2(n²+2n+1)=0
(n+1)²=0/2
:.n=-1
.
x - 3x^2 = 5+ 2x - x^2
0=5+ 2x - x^2-x +3x^2
0=5+x+2x²
2x²+x+5=0
comparing above equation with ax²+bx +c we get
a=2
b=1
c=5
x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1
={-1±√-39}/2
You might be interested in
The value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.
<h3>What is the value of sin (π/2)?</h3>The value of sin (π/2) is equal to the number 1. The value of the sin-1(1) has to be find out.
Suppose the value of this function is <em>x</em>. Thus,
Solve it further,
......1
The value of sin (π/2) and -sin (-π/2) is equal to 1 such that,
Put this value in the equation 1,
Thus, the range will be,
Thus, the value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.
Learn more about the sine values here;