1.

-5 is one of the zeros of 2x²+px-15 , zeros of p(x²+x)+k are equal to each other . Find the value of k

Answer»

ong>GIVEN :-

  • -5 is one of the zeroes of POLYNOMIAL 2x² + px - 15.
  • Zeroes of p(x² + x) + k are equal.

TO FIND :-

  • Value of 'k'.

SOLUTION :-

Part 1 :-

-5 is one of the zeroes of 2x² + px - 15.

So, for x= -5 , 2x² + px - 15 = 0

→ 2(-5)² + p(-5) - 15 = 0

→ 2(25) - 5P - 15 = 0

→ 50 - 5p - 15 = 0

→ 35 - 5p = 0

→ 5p = 35

→ p = 35/5

→ p = 7

Hence , value of p is 7.

Part 2 :-

Zeroes of [ p(x² + x) + k ] are equal.

Putting value of p,

[ 7(x² + x) + k ]

[ 7x² + 7X + k ] is the required equation.

For equal zeroes , discriminant = 0.

★ D = (b² - 4ac) = 0

Here ,

  • a → Coefficient of x².
  • b → Coefficient of x.
  • c → Constant.

We have ,

  • a = 7
  • b = 7
  • c = k

Putting values,

→ 7² - 4(7)(k) = 0

→ 49 - 28k = 0

→ 28k = 49

→ k = 49/28

→ k = 7/4

Hence , value of k is 7/4.


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