Question

(k+1)x^2-2kx+1=0
Find the value of k it has real and equal roots

Answer 1

Answer:

The given equation is ( k + 1 ) x² - 2 k x + 1 =  0

Comparing with ax² + bx  + c = 0 we get :

a = k + 1

b = - 2 k

c = 1

If the equation has equal roots then :

b² = 4 ac

⇒ ( - 2 k )² = 4 ( k + 1 ) ( 1 )

⇒ 4 k² = 4 ( k + 1 )

⇒ k² = k + 1

⇒ k² - k - 1 = 0

Using the quadratic formula we get :

$k=\dfrac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}\\\\\implies k=\dfrac{1\pm \sqrt{1+4}}{2}\\\\\implies k=\dfrac{1\pm \sqrt{5}}{2}\\\\\implies k=\dfrac{1\pm2.236}{2}\\\\\mathsf{Either\:k=\dfrac{1+2.236}{2}\implies \dfrac{3.236}{2}\implies1.618 }\\\\\mathsf{Or\:k=\dfrac{1-2.236}{2}\implies \dfrac{-1.236}{2}\implies-0.618 }$

The value is either - 0.618 or 1.618 .

MORE INFO :

If the solution has unequal , real roots then apply :

b² > 4 ac .

Otherwise in case of unreal roots :

b² < 4 ac

The general form of quadratic equation is a x² + b x + c = 0 .

The quadratic equation is an equation of degree 2 .

Answer 2

Given equation is ( k + 1)x^2 - 2kx + 1 = 0

From the properties of quadratic equations we know that if an equation has real and equal roots, numeric value of discriminant of that equation is equal to 0.

Therefore, discriminant of ( k + 1 )x^2 - 2kx + 1 is equal to 0.

On comparing the equation ( k + 1 )x^2 - 2kx + 1 = 0 with ax^2 + bx + c = 0 we get

a = ( k + 1 )

b = - 2k

c = 1

We know, discriminant = b^2 - 4ac

In this case,where equation has real and equal roots : -

= > Discriminant = 0

= > b^2 - 4ac = 0

= > ( - 2k )^2 - 4[ ( k + 1 ) x 1 ] = 0

= > 4k^2 - 4( k + 1 ) = 0

= > 4k^2 - 4k - 4 = 0

= > k^2 - k - 1 = 0

Now, comparing this equation with ax^2 + bx + c = 0, we get

a = 1, b = - 1, c = - 1

By Quadratic Formula : -

$\box{ Quadratic\: Formula, x = \dfrac{ -b\pm \sqrt{b^2-4ac}}{2a}}$

Thus,

$\implies k = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4(1 \times - 1) } }{2 \times 1} \\ \\ \implies k = \dfrac{1 \pm \sqrt{ {1}^{2} + 4} }{2}$

Hence the required value of k ja ( 1 ±√5) / 2