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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

If a, b are real then show that the roots of the equation (a – b)x2 – 6(a + b)x – 9(a – b) = 0 are real and unequal - Mathematics

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Question

If a, b are real then show that the roots of the equation (a – b)x2 – 6(a + b)x – 9(a – b) = 0 are real and unequal

Sum

Solution

(a – b)x2 – 6(a + b)x – 9(a – b) = 0

Δ = b2 – 4ac

= (– 6(a + b)2 – 4(a – b)(– 9(a – b))

= 36(a + b)2 + 36(a – b)2

= 36(a2 + 2ab + b2) + 36(a2 – 2ab + b2)

= 72a2 + 12b2

= 72(a2 + b2) > 0

∴ The roots are real and unequal.

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Quadratic Equations
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Chapter 3: Algebra - Exercise 3.13 [Page 119]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 3 Algebra
Exercise 3.13 | Q 4 | Page 119
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