Question

Find the least positive value of k for which the equationX²+KX+4=0 has real roots.

Answer 1

Answer:

The least positive value of k for which the quadratic has real roots is 4 .

Step-by-step explanation:

Given as :

The quadratic equation is

x² + k x + 4 = 0

The equation is in form of

ax² + b x + c = 0

According to question

∵ The equation has real roots

For real and equal roots , discriminant = 0

i.e D = 0

So, b² - 4 a c = 0

Or,  k² - 4 × 1 × 4 = 0

Or,  k² - 16 = 0

Or,  k² = 16

∴    k = √16

i.e  k = 4 , - 4                      ...........1

Again

For real and unequal roots , discriminant $>$ 0

i.e D $>$ 0

So, b² - 4 a c $>$ 0

Or,  k² - 4 × 1 × 4 $>$ 0

Or,  k² - 16 $>$ 0

Or,  k² $>$ 16

∴    k $>$ √16

i.e k $>$ 4 , - 4                   ...........2

Now, From 1 and 2

As the roots are always real , and positive

So, The least positive value of k = 4

Hence, The least positive value of k for which the quadratic has real roots is 4 . Answer

Answer 2

Given:-

• the equation x²+kx + 4 = 0 has real roots.

To find:-

• Find the value of k.?

Solutions:-

• The given quadratic equation is x²+kx + 4 = 0 and roots are real.

Here,

• a = 1
• b = k
• c = 4

We know that;

D => b² - 4ac

Putting the value of a = 1, b = k and c = 4

• D => b² - 4ac

=> (k)² - 4 × 1 × 4

=> k² - 16

The given equation will have real and equal root,

• D = 0

=> k² - 16 = 0

Now factroizing of the above equation.

=> k² - 16 = 0

=> k² = 16

=> k = √16

=> k = +,- 4

Now, the value of k is positive.

Hence, the value of k is 4.