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The discriminant of a quadratic and real solutions

  • Thread starter Keen94
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  • #1
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Homework Statement


Given a series of mathematical statements, some of which are true and some of which are false. Prove or Disprove:
1. A sufficient condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac>5.
2.A necessary condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac=0.

Homework Equations


x(px→qx)


The Attempt at a Solution


1.[/B]A sufficient condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac>5.

If b2-4ac>5, then ax2+bx+c=0 (a≠0) has a real root.
ax2+bx+c=0
x2+(b/a)x+(c/a)=0
x2+(b/a)x+(b2/4a2)+(c/a)=(b2/4a2)
(x+(b/2a))2+(c/a)=(b2/4a2)
(x+(b/2a))2=(b2/4a2)-(c/a)
(x+(b/2a))2=(b2/4a2)-(4ac/4a2)
x+(b/2a)=(±√b2-4ac)/2a
x=(-b±√b2-4ac)/2a
±√b2-4ac≥0
b2-4ac≥0
Since 5>0, b2-4ac>5 is sufficient condition for the equation to a have real root.

2. A necessary condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac=0.

If and only if b2-4ac=0, then ax2+bx+c=0 (a≠0) has a real root.
From 1. we know that ax2+bx+c=0 (a≠0) has a real root when b2-4ac>5.
Therefore it is false that the equation has a real root if and only if the discriminant is zero.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Given a series of mathematical statements, some of which are true and some of which are false. Prove or Disprove:
1. A sufficient condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac>5.
2.A necessary condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac=0.

Homework Equations


x(px→qx)


The Attempt at a Solution


1.[/B]A sufficient condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac>5.

If b2-4ac>5, then ax2+bx+c=0 (a≠0) has a real root.
ax2+bx+c=0
x2+(b/a)x+(c/a)=0
x2+(b/a)x+(b2/4a2)+(c/a)=(b2/4a2)
(x+(b/2a))2+(c/a)=(b2/4a2)
(x+(b/2a))2=(b2/4a2)-(c/a)
(x+(b/2a))2=(b2/4a2)-(4ac/4a2)
x+(b/2a)=(±√b2-4ac)/2a
x=(-b±√b2-4ac)/2a
±√b2-4ac≥0
b2-4ac≥0
Since 5>0, b2-4ac>5 is sufficient condition for the equation to a have real root.

2. A necessary condition that ax2+bx+c=0 (a≠0) have a real root is that b2-4ac=0.

If and only if b2-4ac=0, then ax2+bx+c=0 (a≠0) has a real root.
From 1. we know that ax2+bx+c=0 (a≠0) has a real root when b2-4ac>5.
Therefore it is false that the equation has a real root if and only if the discriminant is zero.
For part 2 you are only asked whether the discriminant being zero is necessary to have a real root, not whether it is also sufficient (which it is). So you should only be concerned with the "only if" statement. I would leave out the red words.
 
  • #3
41
1
Gotcha', thanks.
 

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