Question

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

p(x2 + x) + k = 0 has equal roots, find the value of k.​

Answer 1

Answer:

k=(-140)

Step-by-step explanation:

x=(-5)

2x²+px-15=0

2(-5)²+p(-5)-15=0

2(25)+(-5)p-15=0

50-5p-15=0

50-15-5p=0

35-5p=0

(-5p)=(-35)

5p=35

p=35/5

p=7

p(x²+x)+k=0

7[(-5)²+(-5)]+k=0

7(25-5)+k=0

7(20)+k=0

140+k=0

k=(-140)

Answer 2

Given:

• -5 is root of 2x² + px - 15 = 0
• p(x²+x) + k has equal root

To find : k

Method & Formula used :

General quadratic equations is represented by - P(x) = ax² + bx + c

Root of equation is given by -

$\sf \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ \: {b}^{2} \: - \: 4ac } \: }{2a}$

• If α is root of quadratic equations , then P(α) = 0

• Also , for equal root values of ( b² -4ac) = 0

Solution :

We know that (-5) is root of p(x) 2x² + px - 15 = 0.

So when we pit x = -5 it's value{ that is P(-5) } will be equal to zero

➝ P(-5) = 2(-5)² + p(-5) - 15

➝ 0 = 2(25) - 5p - 15

➝ 5p = 50 - 15

➝ 5p = 35

➝ p = 35/5

➝ p = 7

_______________________________________________

• p(x²+x) + k has equal root

First of all we need to simplify it

➝ px² + px + k

on putting value of p , we get;

➝ 7x² + 7x + k has equal root.

On comparing with general representation of quadratic equations,

a = 7

b = 7

c = k

For , equal root values of (b²-4ac ) must be zero

$\sf \implies \: {7}^{2} \: - \: 4(7)(k) \: = \: 0 \\ \\ \sf \implies \:49 \: - \: 28k \: = 0 \\ \\ \sf \implies \:28k \: = \: 49 \\ \\ \sf \implies \:k \: = \: \dfrac{49}{28} \\ \\ \sf \implies \:k \: = \dfrac{7}{4}$

_______________________________________________

ANSWER :

k = (7/4)