Question

2. Find the value of k for which the quadratic equation (k+4) x2 + (k+1)x +1 =0 has equal roots

its urgent pls help​

Answer 1

Answer :

The required value of k is 5 (or) -3

Step-by-step explanation :

Given :

the quadratic equation (k+4)x² + (k+1)x + 1 =0 has equal roots

To find :

the value of k

Solution :

For the quadratic equation ax² + bx + c = 0 ;  the nature of roots is determined by the value of discriminant which is given by

★ D = b² - 4ac

So, for the given quadratic equation,

• a = (k + 4)

• b = (k + 1)

• c = 1

Finding the value of discriminant,

➙ D = (k + 1)² - [ 4(k + 4)(1) ]

➙ D = k² + 1² + 2(k)(1) - [ 4k + 16 ]

➙ D = k² + 1 + 2k - 4k - 16

➙ D = k² - 2k - 15

The nature of roots of the given quadratic equation is equal roots.

Then, D = 0

➙ k² - 2k - 15 = 0

➙ k² + 3k - 5k - 15 = 0

➙ k(k + 3) - 5(k + 3) = 0

➙ (k + 3) (k - 5) = 0

⇒ k + 3 = 0 ; k = -3

⇒ k - 5 = 0 ; k = 5

➙ k = 5, -3

Therefore, the value of k can 5 or -3.

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Know more :

If D > 0 ; the roots are real and unequal

If D = 0 ; the roots are real and equal

If D < 0 ; the roots are not real i.e., complex roots

Answer 2

$\bf{Answer}$

Step by step explanation:-

Given :-

(k+4) x² + (k+1)x +1 =0 has equal roots

To find :-

Value of x

To know :-

If the Quadratic equation has equal roots then Discriminant is equal to 0

So, b² -4ac =0

Solution:-

b² -4ac = 0

a = k+4

b = k+1

c = 1

b² -4ac =0

(k+1)² -4(k+4)(1) =0

k² +2k+1 -4k -16 =0

k² -2k -15 = 0

k² +3k -5k -15 =0

k(k+3) -5(k+3) =0

(k + 3)(k -5) = 0

k = -3,5

So, value of K can be -3 or 5

Know more about Quadratic:-

Quadratic equation has two roots

Its general form is ax²+bx + c

Its degree is 2

Its nature of roots can be determined by value of Discriminant

What is Discriminant??

In ax²+bx+c =0

Discriminant is given by b²-4ac

Discriminant is denoted by D

Nature of roots :-

If D =0 roots are real &equal

D >0 roots are real &distinct

D<0 roots are complex and conjugate to each other

D >0 &D is perfect square Roots are rational & distinct

D<0 &D is not perfect square Roots are irrational &conjugate to each other

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Hope this information helps to u &my answer also helps to u

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