Question

Find the value of k for which the equation $x^{2}$+ 4kx + 25 = 0 has no real roots.

Answer 1

ANSWER

K > 5 / 2

Find

value of k for which equation has no real roots

EQUATION:-

x^2 + 4kx + 25 = 0

equation has no real roots

D < 0

B^2 -4AC < 0

(4k) ^2 - 4 (1) (25) < p

16k^2 - 100 < 0

4k^2 - 25 < 0

k^2 > 25 / 4

k > 5 / 2

Answer 2

Given:-

• x² + 4kx + 25 = 0 has no real roots

To find:-

• Value of k

Solution:-

We have,

★ x² + 4kx + 25 = 0

Comparing this with general form of quadratic

equations ax² + bx + c = 0, we get:-

• a = 1
• b = 4k
• c = 25

Now, for no real roots, i.e. imaginary roots,

D ( Discriminant) < 0

b² - 4ac < 0

(4k)² - 4 × 1 × 25 < 0

(4k)² - 100 < 0

(4k)² - (10)² < 0

Since, x² - a² < 0 = - a < x < a

- 10 < 4k < 10

Or $\sf \dfrac{-10}{4}$ < k < $\sf \dfrac{10}{4}$

Hence value of k > $\bf - \dfrac{5}{2}$ or k < $\bf \dfrac{5}{2}$

$\underline {\rule{216}{2}}$