if a,b,c,d belongs to R show that the roots of the equation (a^2+c^2)x^2+2(ab+cd)x+(b^2+d^2)=0 cannot be real unless they are equal
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Answer:
Hope it helps @deba65..
Show that if the roots of the equation (a^2 +b^2) x^2+2x (AC +bd) + c^2 +d^2=0 are real, they will be equal?
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Find the discriminant of the quadratic equation.
The general form of a quadratic equation is:
Ax2+Bx+C=0,A≠0
The discriminant of the above quadratic equation is given by:
Δ=B2−4AC
Why is it called a discriminant?
This value discriminates (distinguishes) the roots of the equation. How?
We know the well-known formula for finding roots of a quadratic equation:
x=−B±B2−4AC−−−−−−−−√2A
The value inside the square root decides the nature of roots of the equation.
If B2−4AC >0 , then the roots are real and distinct (unequal) [Because of the presence of ± ].
If B2−4AC= 0 , then the roots are real and equal [Because square root of 0 is 0].
If B2−4AC< 0 , then the roots are imaginary [Because square root of a negative number is complex or imaginary].
Now, let's find the discriminant.
From the given equation
(a2+b2)x2+2(ac+bd)x+(c2+d2)=0 ,
A=a2+b2
B=2(ac+bd)
C=c2+d2
Δ=B2−4AC= (2(ac+bd))2−4(a2+b2)(c2+d2)
=4(a2c2+b2d2+2acbd)−4(a2c2+b2c2+a2d2+b2d2)
[Since,(a+b)2=a2+b2+2ab]
=4a2c2+4b2d2+8acbd−4a2c2−4b2c2−4a2d2−4b2d2
=8abcd−4b2c2−4a2d2
Here, we can't decide anything since we don't know the nature or values of a,b,c and d.
Given that the roots are real,
Δ≥0
[We don't know whether roots are equal or not. But, since the roots are given to be real, we have the discriminant to be either greater than zero or equal to zero.]
8abcd−4b2c2−4a2d2≥0
−4(b2c2+a2d2−2abcd)≥0
Dividing by -4,
b2c2+a2d2−2abcd≤0
[Multiplying or dividing by a negative number reverses the relational sign. For example, 4 > 3 implies -4 < -3.]
(bc)2+(ad)2−2(bc)(ad)≤0
(bc−ad)2≤0
[Since,a2+b2+2ab=(a+b)2]
Since the roots are real, discriminant can't be less than 0. Therefore,
Δ=(bc−ad)2=0
Thus, the roots are equal.
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Step-by-step explanation:
The general form of a quadratic equation is:
Ax2+Bx+C=0,A≠0
The discriminant of the above quadratic equation is given by:
Δ=B2−4AC
Why is it called a discriminant?
This value discriminates (distinguishes) the roots of the equation. How?
We know the well-known formula for finding roots of a quadratic equation:
x=−B±B2−4AC−−−−−−−−√2A
The value inside the square root decides the nature of roots of the equation.
If B2−4AC >0 , then the roots are real and distinct (unequal) [Because of the presence of ± ].
If B2−4AC= 0 , then the roots are real and equal [Because square root of 0 is 0].
If B2−4AC< 0 , then the roots are imaginary [Because square root of a negative number is complex or imaginary].
Now, let's find the discriminant.
From the given equation
(a2+b2)x2+2(ac+bd)x+(c2+d2)=0 ,
A=a2+b2
B=2(ac+bd)
C=c2+d2
Δ=B2−4AC= (2(ac+bd))2−4(a2+b2)(c2+d2)
=4(a2c2+b2d2+2acbd)−4(a2c2+b2c2+a2d2+b2d2)
[Since,(a+b)2=a2+b2+2ab]
=4a2c2+4b2d2+8acbd−4a2c2−4b2c2−4a2d2−4b2d2
=8abcd−4b2c2−4a2d2
Here, we can't decide anything since we don't know the nature or values of a,b,c and d.
Given that the roots are real,
I hope this answer is helpful for you