Question

If - 5 is root
of the quadratic equation 2x^2+Px-15 = 0
and the quadratic equation p(x^2+x)+k=0 has equal
roots, find the value of
k .​

Answer 1

9514 1404 393

Answer:

  k = 7/4

Step-by-step explanation:

In quadratic ax^2+bx+c, the product of the roots is c/a, and their sum is -b/a. We know that one root of 2x^2+px-15 is -5, so the other one is ...

  (-15/2)/(-5) = 3/2

Then the value of b (p) is ...

  -p/2 = (3/2)+(-5) = -7/2

  p = 7

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Moving to the next part of the question, the roots will have equal values if the equation is a multiple of a perfect square. The constant in the squared binomial is half the x-coefficient.

  7(x^2 +x) +k = 7(x +1/2)^2 = 7x^2+7x +7/4

Comparing terms, we see that k = 7/4.

Answer 2

Answer: the value of k = 1.75

Step-by-step explanation:

-5 is a root of quadratic equation 2x² + px - 15 = 0

so, 2(-5)² + p(-5) - 15 = 0

=> 50 - 5p - 15 = 0

=> 35 - 5p = 0

p = 7

now, put p = 7 in second quadratic equation,

p(x² + x ) + k = 0

=> 7(x² + x ) + k = 0

=> 7x² + 7x + k = 0

above equation has equal roots

so, D = b² - 4ac = 0

=> 7² - 4 × 7 × k = 0

=> 7 - 4k = 0

=> k = 7/4 = 1.75