If ad does not equal bc then prove that equation (a^2 b^2)x^2 (ac b...
Proof:
To prove that the equation has no real roots, we can consider the discriminant of the quadratic equation. The discriminant is given by the expression b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation.
In this case, the quadratic equation is (a^2 b^2)x^2 (ac bd)x (c^2 d^2 )=0.
Discriminant:
The discriminant of the given quadratic equation is (ac bd)^2 - 4(a^2 b^2)(c^2 d^2 ).
Case 1: ad ≠ bc
If ad ≠ bc, then the discriminant is given by (ac bd)^2 - 4(a^2 b^2)(c^2 d^2 ).
Subcase 1.1: ac bd ≠ 0
If ac bd ≠ 0, then the discriminant is (ac bd)^2 - 4(a^2 b^2)(c^2 d^2 ).
Subcase 1.1.1: (ac bd)^2 > 4(a^2 b^2)(c^2 d^2 )
If (ac bd)^2 > 4(a^2 b^2)(c^2 d^2 ), then the discriminant is positive.
In this case, the quadratic equation has two distinct real roots.
Subcase 1.1.2: (ac bd)^2 < 4(a^2="" b^2)(c^2="" d^2="" />
If (ac bd)^2 < 4(a^2="" b^2)(c^2="" d^2="" ),="" then="" the="" discriminant="" is="" />
In this case, the quadratic equation has two complex roots, which means it has no real roots.
Subcase 1.2: ac bd = 0
If ac bd = 0, then the discriminant is 0 - 4(a^2 b^2)(c^2 d^2 ).
In this case, the discriminant is negative or zero, which means the quadratic equation has no real roots.
Case 2: ad = bc
If ad = bc, then the discriminant is (ac bd)^2 - 4(a^2 b^2)(c^2 d^2 ).
Subcase 2.1: ac bd ≠ 0
If ac bd ≠ 0, then the discriminant is (ac bd)^2 - 4(a^2 b^2)(c^2 d^2 ).
In this case, the discriminant is zero, which means the quadratic equation has one real root.
Subcase 2.2: ac bd = 0
If ac bd = 0, then the discriminant is 0 - 4(a^2 b^2)(c^2 d^2 ).
In this case, the discriminant is negative or zero, which means the quadratic equation has no real roots.
Conclusion:
Therefore, regardless of whether ad ≠ bc or ad = bc, the quadratic equation (a^2 b^2)x^2 (ac bd)x
If ad does not equal bc then prove that equation (a^2 b^2)x^2 (ac b...
(a²+b²)x²+2(ac+bd)x+(c²+d²)=0
a=(a²+b²) b=2(ac+bd) c=(c²+d²)
putting a,b,c in formula
=> [-b +/- √(b²-4ac)]/2a
=> {-2(ac+bd) +/-√[4(ac+bd)²-4(a²+b²)(c²+d²)]}/2(a²+b²)
=> [-2(ac+bd)+/- √4a²c²+4b²d²+8abcd-4a²c²-4a²d²-4b²c²-4
b²d²]/2(a²+b²)
=> -2(ac+bd) +/- √ - (4a²d²-8abcd+4b²c²)
Hence root cant be negative.
So, it has no real roots.
That's all🙂
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