Question

If p and q are the zeroes of 7x2

-49x +(5k+4). Also 4p +3q =24,

then find the value of k.​

Answer 1

Answer:

k = 1.6

Step-by-step explanation:

Given polynomial: 7x² - 49x + ( 5k + 4 ) = 0

We know that in a quadratic ax² + bx + c = 0, the sum of zeroes is -b/a

So, sum of zeroes in given the quadratic = -b/a = - (-49)/7 = 49/7 = 7

                  i.e., p + q = 7 ⇒ 3 ( p + q ) = 3 ( 7 ) ⇒ 3p + 3q = 21

Given, 4p + 3q = 24

⇒ p + 3p + 3q = 24

⇒ p  + 21 = 24

⇒ p = 24 - 21

⇒ p = 3

p + q = 7 ⇒ 3 + q = 7 ⇒ q = 7 - 3 ⇒ q = 4

We know that in the quadratic ax² + bx + c = 0, product of roots = c/a

So, product of roots in given polynomial = c/a = (5k + 4)/7

                     i.e., p x q = 5k + 4

⇒ 3 x 4 = 5k + 4

⇒ 12 = 5k + 4

⇒ 5k = 12 - 4

⇒ 5k = 8

⇒ k = 8/5 = 1.6

⇒ k = 1.6

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