Question

If sum of the zeroes of the polynomial x²-(k+3)x+5k-3 is equal to one fourth of the product of the zeroes, find the value of k​

Answer 1

Solution :

Let p(x) = x² - ( k + 3 )x + ( 5k-3)

Compare p(x) with ax² + bx + c , we

get

a = 1 , b = -( k + 3 ), c = 5k-3

i ) sum of the roots = -b/a

= - [-(k+3) ]/2

= ( k+3 )/2 ----( 1 )

ii ) Product of the roots = c/a

= ( 5k - 3 ) ----( 2 )

According to the problem given,

( 1 ) = ( 1/4 ) of ( 2 )

=> ( k + 3 )/2 = ( 5k - 3 )/4

=> 4( k + 3 ) = 5k - 3

=> 4k + 12 = 5k - 3

=> 4k - 5k = -3 - 12

=> - k = -15

=> k = 15

Therefore,

k = 15

Answer 2

Step-by-step explanation:

Given :

p(x) =  x² - (k + 3)x + (5k - 3).

Sum of zeroes of p(x) = 1/4rth of Product of zeroes.

To find :

Value of k ?

Solution :

We have,

➝  x² - (k + 3)x + (5k - 3).

Comparing it with ax² + bx + c we get,

• a = (1)
• b = -(k + 3)
• c = (5k - 3)

Sum of roots, (-b/a)

➝ [-(k + 3)]/2

➝ (k + 3)/2 -- (1)

Product of roots, (c/a)

➝ (5k - 3) -- (2)

According to the question,

➝ Sum of zeroes = 1/4rth of Product of zeroes

Substituting we get,

• (k + 3)/2 = (5k - 3)/4
• 4(k + 3) = 5k - 3
• 4k + 12 = 5k - 3
• 4k - 5k = -3 - 12
• -k = -15
• k = 15

∴ Value of k is 15.

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