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Let $a$ be a constant. If quadratic equation $(ax-1)^2+a^2 -a-2 = 0$ has equal roots, then $a=$?

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    Try 'Googling' "when does a quadratic equation have equal roots". – 2017-01-05
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    @copper.hat: did _you_ try googling it? I did, and all I got was your unhelpful comment! – 2017-01-05
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    @TonyK: Of course I did. The top of the (my) display shows all of the relevant information, discriminant, roots, description. So, for me, the search would have been helpful, not recursive. Actually, I just tried the same search from three different world locations and each returns the same first page. On a mobile search, the first thing returned states (paraphrasing slightly) "a quadratic equation has equal roots if and only if discriminant is zero". – 2017-01-05

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Hint:

$$(ax-1)^2 = -a^2+a+2$$

It has equal root when $-a^2+a+2=0$.

$$(-a+2)(a+1)=0$$

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    they are all equal. I just move the non-quadratic part of your original question to the other side of the equation. – 2017-01-05
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Hint:

  • A quadratic equation has equal roots iff its discriminant is zero.

  • A quadratic equation has equal roots iff these roots are both equal to the root of the derivative.

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    The second hint is easier to apply in this case. – 2017-01-05
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Your equation can be written as $$a^2x^2-2ax+a^2−a−1=0$$ Just equalize the Discriminant with $0$ i.e. in equation $ax^2+bx+c$ the roots will be equal if $$D=b^2-4ac=0$$.

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    Regarding the joke you posted: the third derivative is sometimes referred to as the "jerk". – 2017-01-06