Question

If the equation x²-4|x|+k=0 has four distinct real roots,then the number of integral values of K is​

Answer 1

Answer:

x2-4IXI+K=6.36

Step-by-step explanation:

2/-15*963%84

Answer 2

Answer:

Intregal value of k is (0,4)

Step-by-step explanation:

The equation is :${x}^{2} - 4 |x| + k = 0$

We know,

${x}^{2} = { |x| }^{2}$

So,${ |x| }^{2} - 4 |x| + k = 0$

There are two key facts to remember here,

For this equation to have four real roots.

1.D>0 and 2.

$|x| > 0$

Let's solve for D here-

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

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

$D = 16 - 4k$

Now$D > 0$

So,$16 - 4k > 0$

$k < 4$

Therefore $|x|$belongs to R.

The first condition is satisfied i.e k belongs to R.

The second condition-

It is important to note here that,for a quadratic equation:$a {x}^{2} + bx + c = 0$with roots $\alpha$and $\beta$,for $\alpha$and $\beta$

to be positive two conditions need to be satisfied-

$\alpha + \beta > 0 \\ - \frac{b}{a} > 0$

and

$\alpha \beta > 0 \\ \frac{c}{a} > 0$

Now,

$- \frac{b}{a} = - \frac{ - 4}{1} = 4 > 0$Which is true.

Also,$\frac{c}{a} = \frac{k}{1} = k > 0$

We get,$k > 0 \: \: and \: k < 4$

So value of k is belongs to (0,4).

Hence intregal value of k is (0,4)