Question

If quadratic equation ax^2-4ax+2a=0 has repeated (equal)roots ,a=?????

Answer 1

OK...
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okk

Answer 2

Hey user !!

Here's the answer you are looking for

Roots of a quadratic equation are

$\frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a}$

&

$\frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

$If \: \sqrt{ {b}^{2} - 4ac} \: \: becomes \: zero \: then \: both \: the \: roots \: will \: be \: equal$

$So, \: \\ \sqrt{ {b}^{2} - 4ac} = 0 \\ \\ {b}^{2} - 4ac = 0 \\ {b}^{2} = 4ac \\ \\ a = \frac{ {b}^{2} }{4c}$

In the question,
for ax² - 4ax + 2a

a = a
b = -4a
C = 2a

So, a = (-4a)²/(4×2a)

a = 16a²/8a

8a² = 16a²

16a² - 8a² = 0

8a² = 0

a² = 0

a = 0

So, for a = 0, the quadratic equation ax² - 4ax + 2a will have equal solutions.

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