Question

The value of k for which the roots of the given equation are real is kx²-3x+1=0​

Answer 1

Given,

• Quadratic polynomial, kx² - 3x + 1 = 0
• Roots are real and equal.

To Find,

• The value of k.

According to question,

Given, quadratic polynomial, kx² - 3x + 1 = 0,

On comparing with, ax² + bx + c = 0 , We get ;

⇒ a = k , b = -3 , c = 1

Given, roots are real and equal,

Discriminate = 0

⇒ b² - 4ac = 0

[ Put the values ]

⇒ (-3)² - 4 × k × 1 = 0

⇒ 9 - 4k = 0

⇒ -4k = -9

⇒ k = 9/4

Therefore,

The value of k is 9/4.

Answer 2

⇒SOLUTION

Given equation: kx²-3x+1=0

We can see that it is in the form of ax²+bx+c=0

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

Now Finding the Discriminant of the equations

⇢D=b²-4ac

⇢D=(-3)²-4×k×1

⇢D=9-4k

For real root we must have D≥0

Note,that in the question the equation has only real roots ,so for the real roots Discriminant can be equal of greater that 0.

Therefore.

⇢9-4k≥0

⇢-4k≥ -9

⇢k≥-9/4

⇢k ≥ 9/4

Hence,k≥9/4 is the value for which the equation has real roots

More Information:-

•When D>0, then the nature of the roots of the equation will be real and unequal and the roots will be ⇁-b∓√D/2a

•When D=0, then the nature of the roots of the equation will be real and equal ⇁-b/2a

•When D<0, then the nature of the roots of the equation will be no real roots ⇁none