Question

Find the value of K for
which the roots of quadratic
equation are real and equal
4x²-3kx+1=0​

Answer 1

Question :

Find the value of k for which the roots of quadratic equation are real and equal 4x²-3kx+1=0

Theory :

For a Quadratic equation of the form

ax²+bx+c= 0 , the expression b²-4ac is called the discriminant.

Nature of roots

The roots of a quadratic equation can be of three types.

1. If D>0, the equation has two distinct real roots.
2. If D=0, the equation has two equal real roots.
3. If D<0, the equation has no real roots.

Solution :

We have , 4x²-3kx+1=0

On comparing with the standard form of Quadratic equation ax²+bx+c= 0.

Here ,

a= 4

b = -3k

and c = 1

For equal and real roots ;

Discrimant = 0

$\sf\:b^2-4ac=0$

$\sf\implies\:b^2=4ac$

$\sf\implies\:(-3k)^2=4\times4\times1$

$\sf\implies\:9k^2=16$

$\sf\implies\:(3k)^2-(4)^2=0$

$\sf\implies\:(3k+4)(3k-4)=0$

$\rm\implies\:k=\dfrac{4}{3}\:or\:\dfrac{-4}{3}$

Therefore , The value of k = -4/3 or 4/3

Answer 2

Answer:

Given :-

• The roots of qradratic equation are real and equal is 4x² - 3kx + 1 = 0

To Find :-

• What is the value of k.

Solution :-

Given equation :

$\mapsto$ 4x² - 3kx + 1 = 0

where, a = 4, b = - 3k, c = 1

Since, the roots are real and equal.

$\leadsto$ The discriminant = 0

$\therefore$ b² - 4ac = 0

=> (- 3k)² - 4.4.1 = 0

=> 9k² - 16 = 0

=> (3k)² - (4)² = 0

=> (3k + 4)(3k - 4) = 0

=> 3k + 4 = 0 ; 3k - 4 = 0

=> 3k = - 4 ; 3k = 4

=> k = $\dfrac{- 4}{3}$ ; k = $\dfrac{4}{3}$

$\Longrightarrow$ k = $\dfrac{- 4}{3}$ , $\dfrac{4}{3}$

$\therefore$ The value of k is $\dfrac{- 4}{3}$ , $\dfrac{4}{3}$

________________________________