If the roots of the quadratic equation 9x² + kx + k = 8 are real and equal, what is the value of k?
Answers
Given:-
If the roots of the quadratic equation 9x² + kx + k = 8 are real and equal, what is the value of k?
Solution:-
9x² + kx + k = 8
=> 9x² + kx + k-8=0
Here , a = 9 , b = k and c = k-8
For real and equal roots :
• D = 0
=> b² - 4 ac = 0
=> k² - 4 (9)(k-8)= 0
=> k² -36k + 288 = 0
=> k² -24k - 12k +288=0
=> k ( k-24 ) -12 (k- 24) =0
=> (k-24)(k-12)=0
=> k = 12 and 24



OR
Solution :-
Solution :-We know that
9x² + kx + k = 8
9x² + kx + k - 8 = 0
D = b² - 4ac
Here
b = k
a = 9
c = k - 8
By putting values
0 = (k)² - 4(9)(k - 8)
0 = k² - 36(k - 8)
0 = k² - 36k + 288
By quadratic formula
x = -b ± √(b² - 4ac)/2a
x = -(-36) ± √(36² - 4(1)(288))/2(1)
x = 36 ± √(1296 - 1152)/2
x = 36 ± √(144)/2
x = 36 ± 12/2
x = 36 + 12/2 or 36 - 12/2
x = 48/2 or 24/2
x = 24 or 12
Thank you
#BeBrainly
Answer:
9x² + kx + k = 8
=> 9x² + kx + k-8=0
Here , a = 9 , b = k and c = k-8
For real and equal roots :
• D = 0
=> b² - 4 ac = 0
=> k² - 4 (9)(k-8)= 0
=> k² -36k + 288 = 0
=> k² -24k - 12k +288=0
=> k ( k-24 ) -12 (k- 24) =0
=> (k-24)(k-12)=0
=> k = 12 and 24