The discriminant of a quadratic and real solutions

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A sufficient condition for the quadratic equation ax² + bx + c = 0 (where a ≠ 0) to have a real root is that the discriminant b² - 4ac > 5, which confirms the existence of real solutions. Conversely, it is false that the condition b² - 4ac = 0 is necessary for the equation to have a real root, as real roots can exist when the discriminant is greater than zero. The discussion clarifies that while a zero discriminant indicates a double root, it does not exclusively determine the presence of real roots. Thus, the discriminant's value must be greater than or equal to zero for real solutions, but it does not have to be exactly zero. Understanding these conditions is crucial for solving quadratic equations effectively.
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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.
 
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Keen94 said:

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.
 
Gotcha', thanks.
 
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