Find the value of k for which the quadratic equation (k - 2)x^2 (2k - 3)x +(5k - 6)=0
has real roots
Answers
Answer :
k € [7/6 , 3/2]
Note:
★ The possible values of the variable which satisfy the equation are called its roots or solutions .
★ A quadratic equation can have atmost two roots .
★ The general form of a quadratic equation is given as ; ax² + bx + c = 0
★ The discriminant , D of the quadratic equation ax² + bx + c = 0 is given by ;
D = b² - 4ac
★ If D = 0 , then the roots are real and equal .
★ If D > 0 , then the roots are real and distinct .
★ If D < 0 , then the roots are unreal (imaginary) .
Solution :
Here ,
The given quadratic equation is ;
(k - 2)x² + (2k - 3)x + (5k - 6) = 0
Comparing the given quadratic equation with the general quadratic equation ax² + bx + c = 0 , we have ;
a = k - 2
b = 2k - 3
c = 5k - 6
Now ,
The discriminant D of the given quadratic equation will be given as ;
=> D = b² - 4ac
=> D = (2k - 3)² - 4•(2k - 3)•(5k - 6)
=> D = (2k - 3)² - 4(10k² - 12k - 15k + 18)
=> D = 4k² - 12k + 9 - 4(10k² - 27k + 18)
=> D = 4k² - 12k + 9 - 40k² + 108k - 72
=> D = -36k² + 96k - 63
Also ,
We know that , for real roots of a quadratic equation , its discriminant must be equal to or greater than zero .
Thus ,
=> D ≥ 0
=> -36k² + 96k - 63 ≥ 0
=> -3(12k² - 32k + 21) ≥ 0
=> 12k² - 32k + 21 ≤ 0
=> 12k² - 14k - 18k + 21 ≤ 0
=> 2k(6k - 7) - 3(6k - 7) ≤ 0
=> (6k - 7)(2k - 3) ≤ 0
=> k € [7/6 , 3/2]
Hence ,
k € [7/6 , 3/2]
