If p,q r real and p≠ q then show that the root of the equation p-qx^2+5(p+q)x -2(p-q)=0 are real and unequal.
Answers
Answered by
253
(p - q)x² + 5(p+q) x - 2(p-q) = 0
a = p - q
b = 5 (p + q)
c = -2(p - q)
As the roots are real and unequal ,
b² - 4ac > 0
i.e.,
{5 ( p + q) }² - 4 ( p - q) ( -2 { p - q} )
25 ( p² + 2pq + q²) + 8 ( p -q) ( p - q)
25 ( p² + 2pq + q²) + 8 ( p - q)²
25 ( p² + 2pq + q²) + 8p² - 16pq + 8q²
25p ² + 50 pq + 25q² + 8p² - 16pq + 8q² > 0
∴ The equation has roots which are real and unequal
a = p - q
b = 5 (p + q)
c = -2(p - q)
As the roots are real and unequal ,
b² - 4ac > 0
i.e.,
{5 ( p + q) }² - 4 ( p - q) ( -2 { p - q} )
25 ( p² + 2pq + q²) + 8 ( p -q) ( p - q)
25 ( p² + 2pq + q²) + 8 ( p - q)²
25 ( p² + 2pq + q²) + 8p² - 16pq + 8q²
25p ² + 50 pq + 25q² + 8p² - 16pq + 8q² > 0
∴ The equation has roots which are real and unequal
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Answered by
66
Answer:
Step-by-step explanation:
Solution :-
The given equation is (p - q)x² + 5(p + q)x - 2(p - q) = 0
Let D be the discriminant of the given equation.
Then,
D = b² - 4ac
= 25(p + q)² - 4(p - q) × [- 2(p - q)
= 25(p + q)² + 8(p -q)²
We find that
25(p + q)² > 0 and 8(p - q)² > 0
Therefore,
D = 25(p + q)² + 8(p - q)² > 0
Hence, roots of the equation are real and unequal.
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