Question

please solve this question............​

Answer 1

$Given \: (-5) \: is \: a \: root \: Quadratic \\equation \: 2x^{2} + px -15 = 0$

/* Replace x by (-5) , we get *)

$\implies 2 \times (-5)^{2} + p(-5) - 15 = 0$

$\implies 50 - 5p - 15 = 0$

$\implies 35 - 5p = 0$

$\implies - 5p = -35$

$\implies p = \frac{-35}{-5}$

$\implies p = 7$

/* Substitute p = 7 and x = -5 , in p(x²+x) + k = 0 , we get */

$7[ (-5)^{2} + (-5) ] + k = 0$

$\implies 7 ( 25 - 5 ) + k = 0$

$\implies 7 \times 20 + k = 0$

$\implies 140 + k = 0$

$\implies k = - 140$

Therefore.,

$\green { p = 7 \: and \: k = -140 }$

•♪••

Answer 2

$\huge{\mathcal{\orange{\underline{k=35/28}}}}$

Given:

-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

To find :

Value of "k"

Explanation :

First we have to find the value of p ,

$2x { }^{2} + px - 15 = 0 \\ 2( - 5) {}^{2} + p( - 5) - 1 5 = 0 \\ 2(25) - 5p - 15 = 0 \\ 50 - 5p - 15 = 0 \\ 35 - 5p = 0 \\ 3p = 35 \\ p = \frac{35}{5} \\ p = 7$

Now putting the value of p in equation :

$p(x {}^{2} + x) + k = 0 \\ px {}^{2} + px + k = 0 \\ 7x {}^{2} + 7x + k = 0$

given that the above equation has equal roots. now the condition for equal roots :

$b {}^{2} - 4ac = 0$

b = 7

a = 7

c = k

$(7) {}^{2} - 4(7)(k) = 0 \\ 35 - 28k = 0 \\ k = \frac{35}{28}$