If the roots of the equation(a2+ b2)x2− 2b(a + c)x + (b2+c2) = 0...
(a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0
Roots are real and equal ∴ D = 0
D = b2 − 4ac = 0
⇒ [−2b(a+c)]2 − 4(a2 + b2)(b2 + c2) = 0
⇒ b2(a2 + c2 + 2ac) −(a2b2 + a2c2 + b4 + c2c2) = 0
⇒ b2a2 + b2c2 + 2acb2 − a2b2 − a2c2 − b4 − b2c2 = 0
⇒ 2acb2 − a2c2 − 2acb2 = 0
⇒ (b2 − ac)2 = 0
⇒ b2 = ac
If the roots of the equation(a2+ b2)x2− 2b(a + c)x + (b2+c2) = 0...
We can rearrange the given equation to get:
(a^2 - b^2)x^2 + 2abx - 3ab = 0
The discriminant of this quadratic equation is:
D = (2ab)^2 - 4(a^2 - b^2)(-3ab)
= 4a^2b^2 + 12a^3b - 12ab^3
We want to find the conditions for which D is negative, which will imply that the quadratic equation has no real roots.
4a^2b^2 + 12a^3b - 12ab^3 < />
Dividing both sides by 4ab, we get:
ab + 3a^2 - 3b^2 < />
We can rewrite this inequality as:
ab < 3b^2="" -="" />
Now, we need to consider two cases:
1. b > 0: In this case, we can divide both sides by b^2 and get:
a/b < 3="" -="" />
Letting x = a/b, we can rewrite this as:
x^2 + 3x - 3 < />
This is a quadratic inequality that has solutions for:
-3 - sqrt(21) < x="" />< -3="" +="" />
Since x = a/b, this implies that:
a/b < -3="" -="" sqrt(21)="" or="" a/b="" /> -3 + sqrt(21)
2. b < 0:="" in="" this="" case,="" we="" can="" divide="" both="" sides="" by="" -b^2="" and="" />
-a/b < 3="" -="" />
Letting x = a/b, we can rewrite this as:
x^2 - 3x - 3 < />
This is a quadratic inequality that has solutions for:
(3 - sqrt(21)) < x="" />< (3="" +="" />
Since x = a/b, this implies that:
a/b < 3="" -="" sqrt(21)="" or="" a/b="" /> 3 + sqrt(21)
Therefore, the roots of the given quadratic equation are real if and only if:
a/b < -3="" -="" sqrt(21)="" or="" a/b="" /> -3 + sqrt(21) or a/b < 3="" -="" sqrt(21)="" or="" a/b="" /> 3 + sqrt(21)