Question

if -5 is a 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

Let the other root be A
therefore,
-5*A=c/a
-5*A= -15/2
A=3/2
now 3/2-5= -p/2
 -7/2= -p/2
p= -7
now as equal roots b^2-4ac=0
7^2-4(7)(k)=0
49-28k=0
49=28k
k=7/4

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Root is -5

2x² + px - 15 = 0 ......(1)

$\textbf{\underline{Equal\;Roots :- }}$

p(x² + x ) + k = 0 ……(2)

${\boxed{\sf\:{Substitute\;value\;of\;x\;in\;(1)}}}$

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

$\tt{\rightarrow p=\dfrac{35}{5}}$

p = 7

${\boxed{\sf\:{Substitute\;value\;of\;p\;in\;(2)}}}$

p(x² + x ) + k = 0

7(x² + x ) + k = 0

7x² + 7x + k = 0

Here we have,

${\boxed{\sf\:{a = 7\;,\;b = 7\;and\;c = k}}}$

D = b² - 4ac

$\textbf{\underline{Quadratic\;equation\;has\;equal\; roots}}$

D = 0

b² - 4ac = 0

7² - 4(7)(k) = 0

49 - 28k = 0

49 = 28k

$\tt{\rightarrow k=\dfrac{49}{28}}$

$\tt{\rightarrow k=\dfrac{7}{4}}$

$\textbf{\underline{Therefore}}$

$\tt{\rightarrow k=\dfrac{7}{4}}$