Question

for what value of k given roots of quadratic equations
$(k + 4){x}^{2} + (k + 1)x + 1 = 0 \: equal$
​

Answer 1

EXPLANATION.

Quadratic equations.

⇒ (k + 4)x² + (k + 1)x + 1 = 0.

As we know that,

⇒ D = Discriminant Or b² - 4ac.

For real and equal roots : D = 0.

⇒ (k + 1)² - 4(k + 4)(1) = 0.

As we know that,

Formula of :

⇒ (x + y)² = x² + y² + 2xy.

Using this formula in equation, we get.

⇒ (k² + 1 + 2k) - 4(k + 4) = 0.

⇒ k² + 1 + 2k - 4k - 16 = 0.

⇒ k² - 2k - 15 = 0.

Factorizes the equation into middle term splits, we get.

⇒ k² - 5k + 3k - 15 = 0.

⇒ k(k - 5) + 3(k - 5) = 0.

⇒ (k + 3)(k - 5) = 0.

⇒ k = -3 and k = 5.

                                                                                                                     

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

Answer:

Given :-

$\bf(k + 4) {x}^{2} + (k + 1)x + 1 = 0$

To Find :-

Value of k

Solution :-

We know that

D = b² - 4ac

And

$\sf \: (a + b {)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Now,

$\sf \: ( {k}^{2} + 1 + 2k)- 4(k - 4)$

$\sf \: ({k}^{2} + 1 + 2k )- 4k - 16$

$\sf \: {k}^{2} - 4k + 2k -16 - 1$

$\sf {k}^{2} - 2k - 15$

Now,

By spilting

$\sf \: {k}^{2} - (5k - 3k) - 15$

$\sf \: {k}^{2} - 5k + 3k - 15$

$\sf \: k(k - 5) + 3(k - 5)$

$\sf \: (k + 3)(k - 5) = 0$

Either

k = 0 - 3

k = -3

Or,

k = 0 + 5

k = 5