Question

Complete set of real values of 'k' so that equation x^2 +{√(k-1)}x +k^2 -4=0 has one root less than 0 and other greater than 0 ,is​

Answer 1

given a quadratic equation have one root less than zero and other root greater than zero, find value of 'k'

Explanation:

1. for a given quadratic equation in 'x' and real numbers 'a', 'b' and 'c'                          $ax^2+bx+c=0$ , the roots are given as ,                                                                

                        $->x_1=\frac{-b+\sqrt{b^2-4ac}}{2a},\ \ x_2= \frac{-b-\sqrt{b^2-4ac}}{2a}$     ----(i)  

    2. given equation:  $x^2 +x{\sqrt{k-1}}+(k^2 -4)=0$              

                      $->a=1\ \ \ \ \ b=\sqrt{k-1}\ \ \ \ \ c=k^2-4$          ------(ii)

    3. equating (ii) and (i) we get,

                         [tex]->x_1=\frac{-\sqrt{k-1} +\sqrt{k-1-4(k^2-4)}}{2} \\\\ ->x_2=\frac{-\sqrt{k-1} -\sqrt{k-1-4(k^2-4)}}{2}\\\\->x_1=\frac{-\sqrt{k-1} +\sqrt{k-4k^2+15}}{2}\ \ \ x_2=\frac{-\sqrt{k-1} -\sqrt{k-4k^2+15}}{2}\ [/tex]                          

     4. since it is given that one root is greater than zero (hence, positive)

          and other less than zero (hence, negative) we have,                                

                          [tex]->x_1.x_2<0\ \ \ \ \ \ (product\ is\ negative) \\ ->\frac{(k-1)-(k-4k^2+15)}{4} <0 \\ -> 4k^2-16<0\\ ->k<\sqrt{4} \\ ->-2      5. this can also be solved using direct formula for product of roots of

         a quadratic equation ,    $x_1.x_2=\frac{c}{a}$  

Answer 2

Complete set of real values of 'k' so that equation x² +{√(k-1)}x +k² -4=0 has one root less than 0 and other greater than 0 ,is​   k ∈ [1 , 2)

Given:

• x² +{√(k-1)}x +k² -4=0 has one root less than 0 and other greater than 0

To Find:

• Set of real values of 'k'

Solution:

• Quadratic Equation ax² + bx + c = 0  where a, b , c are real and a≠0
• Product of roots = c/a
• Sum of roots = -b/a
• D = b² - 4ac
• D >0 for Real and distinct roots
• Square root of a negative number is not real

Step 1:

Comparing x² +{√(k-1)}x +k² -4=0   with ax² + bx + c = 0

a = 1  , b =√(k-1) ,   c = k² -4

Step 2:

D > 0  for Real and distinct roots

=> (√(k-1))² - 4(1)(k² - 4) > 0

=> k - 1  - 4k² + 16 > 0

=> -4k² + k  + 15  > 0

=> -1.82 < k < 2.07

Step 3:

√(k-1)  to be real  k - 1 ≥ 0

=> k ≥ 1

Step 4:

As one root less than 0 and other greater than 0 hence product must be less than 0

=> (k² -4)/1 < 0

=> (k + 2)(k - 2) < 0

=> -2 < k  < 2

Step 5:

Find Common Solution for all 3

-1.82 < k < 2.07 ,  k ≥ 1  and   -2 < k  < 2

1 ≤ k  < 2

k ∈ [1 , 2)

Complete set of real values of 'k' so that equation x² +{√(k-1)}x +k² -4=0 has one root less than 0 and other greater than 0 ,is​   k ∈ [1 , 2)