Question

The quadratic equation x² - Kx + 4 =0 : k>0 has both real and distinct roots if

Answer 1

Answer:

compare x² + 4x + k = 0 with ax² + bx + c = 0

a = 1 , b = 4 , c = k

it is given that roots are distinct and real ,

discreaminant ≥ 0

b² - 4ac ≥ 0

4² - 4 × 1 × k ≥ 0

- 4k ≥ - 16

4k ≤ 16

k ≤ 16/4

k ≤ 4

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Step-by-step explanation:

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Answer 2

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}}$

The given Quadratic Equation is

$x² - Kx + 4 =0$

Comparing the given equation with

$a {x}^{2} + bx + c = 0$

We get

$a = 1 \: , b = - k \: , c = 4$

Now the Discriminant

$= {b}^{2} - 4ac$

$= {( - k)}^{2} - 4 \times 1 \times 4$

$= {k}^{2} - 16$

Now the both real and distinct roots if

${b}^{2} - 4ac </strong><strong>></strong><strong> </strong><strong>0$

$\implies \: {k}^{2} - 16 </strong><strong>></strong><strong> 0$

$\implies \: {k}^{2} </strong><strong>></strong><strong> 16$

So k > 4 or k < - 4

Since k > 0

So k<-4 is not possible

Hence k>4

Therefore The quadratic equation x² - Kx + 4 =0 has both real and distinct roots if k>4

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