Question

Find the value of k if the following quadratic equation has equal roots.

Answer 1

EXPLANATION.

Quadratic equation.

⇒ k²x² - 2(2k - 1)x + 4 = 0.

As we know that,

⇒ D = Discriminant Or b² - 4ac.

For real and equal roots : D = 0.

⇒ [-2(2k - 1)]² - 4(k²)(4) = 0.

⇒ [4(2k - 1)²] - 16k² = 0.

⇒ 4(4k² + 1 - 4k) - 16k² = 0.

⇒ 16k² + 4 - 16k - 16k² = 0.

⇒ 4 - 16k = 0.

⇒ 4 = 16k.

⇒ k = 4/16.

⇒ k = 1/4.

                                                                                                                         

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

Step-by-step explanation:

Given:-

$\tt{ {k}^{2} {x}^{2} - 2(2k - 1)x + 4 = 0}$

To find:-

$\tt{Find \: the \: value \: of \: k \: if \: the \: following \: quadratic \: equation \: has \: equal \: roots. }$

Solution:-

Let's solve the problem

We, have

$\tt{ {k}^{2} {x}^{2} - 2(2k - 1)x + 4 = 0}$

Here, a = k²

b = -2(2k-1)

c = 4

It is given that roots are real and equal.

∴ b² - 4ac = 0

According to the question

D = 0

Now,

$\tt{[ - 2(2k - 1){ ]}^{2} - 4 {k}^{2} \times 4 = 0}$

$\tt{4(4 {k}^{2} - 4k + 1) - 16 {k}^{2} = 0}$

$\tt{ \cancel{16 {k}^{2}} - 16k + 4 - \cancel{16 {k}^{2} } = 0}$

$\tt{ - 16k + 4 = 0}$

$\tt{ \cancel{ - }16k = \cancel{- }4}$

$\tt{16k = 4}$

$\tt{k = \cancel{ \frac{4^{1} }{16^{4}} }}$

$\tt{k = \frac{1}{4} }$

Answer:-

$\tt{The \: value \: of \: K \: is \: \: \frac{1}{4} }.$