Let the successive integral multiples of 5 be a, a + 5.⇒ a(a + 5) = 300⇒ a2 + 5a – 300 = 0⇒ a2 + 20a – 15a – 300 = 0⇒ a(a + 20) – 15(a + 20) = 0⇒ (a + 20)(a - 15) = 0⇒ a = - 20, 15Numbers are, -20, -15 or 15, 20

Q. 135.0( 4 Votes )

The product of tw

Class 10th RD Sharma - Mathematics 8. Quadratic Equations

Answer :

Let the successive integral multiples of 5 be a, a + 5.

⇒ a(a + 5) = 300

⇒ a² + 5a – 300 = 0

⇒ a² + 20a – 15a – 300 = 0

⇒ a(a + 20) – 15(a + 20) = 0

⇒ (a + 20)(a - 15) = 0

⇒ a = - 20, 15

Numbers are, -20, -15 or 15, 20

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