for what value of a will (2a - 3)x²+7x-2a=0?
Answers
Correct Question :- For what value of a will (2a - 3)x²+7x-2a = 0 have both equal roots ?
Concept used :-
If A•x² + B•x + C = 0 ,is any quadratic equation,
then its discriminant is given by;
D = B² - 4•A•C
• If D = 0 , then the given quadratic equation has real and equal roots.
• If D > 0 , then the given quadratic equation has real and distinct roots.
• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...
Solution :-
comparing the given expression (2a - 3)x² + 7x - 2a = 0 with A•x² + B•x + C = 0, we get,
- A = (2a - 3)
- B = 7
- C = (-2a)
Given that, the expression has both equal roots.
So,
→ D = 0
→ B² - 4•A•C = 0
Putting all values we get,
→ (7)² - 4*(2a - 3)(-2a) = 0
→ 49 + 8a(2a - 3) = 0
→ 49 + 16a² - 24a = 0
→ 16a² - 24a + 49 = 0
Now, we know that, Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,
→ x = [ -b±√(b²-4ac) / 2a ]
So, again, here we have :-
- a = 16
- b = (-24)
- c = 49 .
- x = a .
Therefore,
→ a = [ -b±√(b²-4ac) / 2a ]
→ a = [-(-24) ± √{(-24)² - 4*49*16} / 2*16 ]
→ a = [24 ± √{576 - 3136} / 32 ]
→ a = [ 24 ± √(-2560) ] / 32
→ a = [ 24 ± {√2560 * √(-1) } ] / 32
→ a = [ 24 ± 16√10i ] / 32
→ a = (24 + 16i√10)/32 or, (24 - 16i√10)/32
→ a = (3 + 2i√10)/4 or, (3 - 2i√10)/4
→ a ∈ {(3 + 2i√10)/4 , (3 - 2i√10)/4} .
Learn more :-
if a^2+ab+b^2=25
b^2+bc+c^2=49
c^2+ca+a^2=64
Then, find the value of
(a+b+c)² - 100 = ...
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