find the value of a such that quadric equation (a-3)x2 +4(a-3)x+4=0
Answers
Answer:
→ (a-3)x² - 2ax + 5a = 0
By using (-b±√b²-4ac)/2a ,
Since roots are positive,
→ [+2a ± √ (4a² - 20a² + 60a)] / 2a > 0
We can't find the answer if we use +ve sign for finding the roots as in inequality we're not allowed to square the numbers on both sides if they are of different signs
Therefore when we use -ve sign,
→ +2a - √(4a² - 20a² + 60a) > 0
(Since 2a × 0 = 0)
→ 2a > √(4a² - 20a² + 60a)
Squaring on both sides
→ 4a² > 4a² - 20a² + 60a
→ 20a² > 60a
Therefore, a > 3
So ‘a' belongs to (3,∞)
Now since we know the range of ‘a', it's time when we determine that the roots are not complex but real and equal/unequal
By using D ≥ 0 , we can find out the range for which the roots will be real and equal/unequal
Since, D = b² - 4ac
→ (2a)² - 4 × 5a × (a-3) ≥ 0
→ 4a² - 20a² + 60a ≥ 0
→ -16a² + 60a ≥ 0
→ 16a² ≤ 60a
→ a ≤ 15/4
Therefore ‘a' should belong to (-∞, 15/4] to be real roots