Question

-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 1

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.

Answer 2

Answer:

Value of K is 7/4

Step-by-step explanation:

Given :

- 5 is one 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

$⟹2 {( - 5)}^{2} + p( - 5) - 15 = 0$

$⟹2(25) - 5p - 15 = 0$

$⟹50 - 5p - 15 = 0$

$⟹35 - 5p = 0$

$⟹5p = 35$

$⟹p = \frac{35}{5}$

$⟹p = 7$

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 - 4 ac) = 0

Hence

A = coefficient of x²

B =coefficient of x

C= constant

We have,

A = 7

B = 7

C = K

Putting values

$➞ {7}^{2} - 4(7)(k) = 0$

$➞49 - 28k = 0$

$➞k = \frac{49}{28}$

$➞k = \frac{7}{4}$

$Hence, \: value \: of \: K \: is \frac{7}{4}$