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If $ax^2+bx+10=0$ does not have two distinct real roots where( $a$ and $b$ are real) then the possible values of $5a+b$ from the given 4 options(as obviously there can be infinite possible values..but I only need to find out which of the four given possible values is one of these ...in short it is a Multi choice-multi-correct question) are-( there can be more than one option correct)

1) -3

2) -2

3) -1

4) 0

my attempt Here are some conclusions I have drawn-

  • it is given that it does not have two distinct real roots so either( both roots are imaginary and conjugates of each other)- $D(b^2-4ac)<0$ (that is as $c=10$ so $b^2-40a<0$...or another case is both roots become equal that is $b^2-40a=0$.

  • also I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible but the answers given are $-2,-1,0$ (Edit:apparently $0$ is not the answer as pointed out be Tim phan below..but I still need help in proving it is equal to $-1,-2$)....also I am unable to get other answers I have tried getting $5a+b$ by keeping $x=5$ to get $5(5a+b+2)$ but I don't know whether it will be equal to zero or less than or greater than zero... As I don't know whether $5$ is a root of $f(x)$ or not..neither it is given in question...

Any help??

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  • $\begingroup$ I think you need more constrains. Here is why. Like you pointed out, essentially, you are solving a system of 2 equations with 3 unknown. $b^2 - 4 a c < 0$ and $b + 5a = N$. $\endgroup$ – Paichu Nov 19 '15 at 3:28
  • $\begingroup$ In the attempted solution, you refer to $b^-40c$. Did you omit saying that $c=10$? Are $a,b,c$ restricted in some way, say to integers? $\endgroup$ – André Nicolas Nov 19 '15 at 3:30
  • $\begingroup$ If that is the case, then you can try going about it in this way. Assume $b + 5a = 0$, then $b = -5a$ and $a = -\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together you will get two equations: $b^2 < -\frac{4}{5}b c$ and $25a^2 < 4a c$. You then proceed with 4 cases $a,b>0$, $a,b<0$, $a>0,b<0$, $a<0,b>0$. If all of them lead to contradiction, then it is not possible. $\endgroup$ – Paichu Nov 19 '15 at 3:42
  • $\begingroup$ Why didn't you write $ax^2+bx+10$, instead of "$c$". Tbh, it looks like you were attempting to make it confusing. $\endgroup$ – YoTengoUnLCD Nov 19 '15 at 3:50
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    $\begingroup$ Let $a=1$ and $b=1.01, 1.02, 1.03, 1.04,\dots$. Don't have distinct real roots, indeed don't have real roots at all. $\endgroup$ – André Nicolas Nov 19 '15 at 4:21
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If you write $5a + b = k$ and use this to get rid of the $a$ in the equation $b^2 - 40a \leq 0$, you get:

$$b^2 - 40\left(\frac{k - b}{5}\right) \leq 0$$

$$b^2 + 8b \leq 8k$$

Complete the square:

$$(b + 4)^2 \leq 8k + 16$$

$$(b + 4)^2 \leq 8(k + 2)$$

Since $b$ is arbitrary, the only restriction is that the left hand side is nonnegative. So you get:

$$0 \leq 8(k + 2)$$

$$-2 \leq k$$

Hence $k = 0, -1, -2$ are all possible.

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The possible values (from those listed) are $0$, $-1$, and $-2$, obtained by $(a,b) = (0,0)$, $(1,-6)$, and $\left(\frac{2}{5},-4\right)$. (The first two are not unique, but the third one is; see below.)

Now, if $ax^2+bx+10$ does not have 2 distinct roots, then the discriminant is nonpositive, i.e. $b^2-40a\le 0$. Hence, $40a\ge b^2$, necessarily implying that $a\ge 0$. Furthermore, $$b^2\le 40a\implies |b|\le\sqrt{40a}\implies b\ge -\sqrt{40a}$$ and hence $5a+b \ge 5a-\sqrt{40a}$. It can be checked that $$\min\limits_{a\ge 0}{\left(5a-\sqrt{40a}\right)} = -2 $$ achieved at $a = \frac{2}{5}$. Hence, if $ax^2+bx+10$ does not have 2 distinct roots, then $$5a+b\ge -2$$ with equality if and only if $a = \frac{2}{5}$ and $b = -\sqrt{40a} = -4$.

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  • $\begingroup$ Why is this reasoning of my wrong....I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible $\endgroup$ – Freelancer Nov 19 '15 at 8:16
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    $\begingroup$ @Freelancer why can't we have $f(3)-f(2) = 0$? For example, if $f$ is constant, then this is true, and there's nothing saying it can't (since if it were constant, it would take the constant value $10$, which is nonzero) $\endgroup$ – Joey Zou Nov 19 '15 at 8:17
  • $\begingroup$ I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..(which can be only possible when both 3,2 are roots are of the equation...which is not allowed (as real roots should not be distinct here) $\endgroup$ – Freelancer Nov 19 '15 at 8:19
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    $\begingroup$ @Freelancer who cares if $f$ doesn't have two distinct roots? It says very little about the difference between its values. For example, $f(x) = x^2+1$ has no real roots, yet $f(1) - f(-1) = 0$, $f(2)-f(-2) = 0$, $f(3) - f(-3)=0$, and so on... $\endgroup$ – Joey Zou Nov 19 '15 at 8:21
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I found that if a = 1/5 and b = -1 then 5a+b = 0. Did i break any rules?

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  • $\begingroup$ You found one solution ! Are there others ? $\endgroup$ – Tom-Tom Nov 19 '15 at 8:35
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Assume $b + 5a = 0$, then $b = -5a$ and $a = -\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together to obtain $b^2 < -\frac{4}{5}b c$ and $25a^2 < 4a c$.

(1) If $a>0, b>0$, then we have: $b<-\frac{4}{5}c$< which implies $c<0$. False.

(2) If $a<0, b<0$, then we have: $-25|a|>4c$ which implies $c<0$. False.

If $a>0, b<0$, you have the same thing as in (1) and if If $a<0, b>0$, you have the same thing as in (2). Thus it is not possible for $b + 5a = 0$. Proceed similarly.


Second part: let $D>0$, assume $5a + b =-D$, so $b = -(D+5a)$. Then for this equation to not have two distinct real roots: $b^2 \leq 40a$. replacing $a$ in to obtain: $ (D+5a)^2 < 40a$ Notice that this implies $a > 0$. Simplify: $$D + 5a < \pm\sqrt{40 a}$$ Let $u = \sqrt{a}$, then we have $5u^2 \mp \sqrt{40}u + D < 0$ is the condition, which is if we can find such $u$ under some condition on $D$, then the same condition is also the condition for $ax^2 + bx + 10 = 0$ to not have two distinct real roots. Solve this to get $$u=\frac{1}{10}\left(\sqrt{40} - \sqrt{40 - 20 D}\right)$$ this has a solution if $D \leq 2$, which means if $D \leq 2$, then the $ax^2 + bx + 10 = 0$ does not have two distinct real roots. Thus the possible solutions are $5a + b = -1, -2$, or (3) and (2). Please check my algebra carefully.

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    $\begingroup$ You can't do that yet, since you do not know if $a$ is positive or negative, which can affect the inequality sign. $\endgroup$ – Paichu Nov 19 '15 at 3:58
  • $\begingroup$ I edit the second part there. Check the algebra carefully though. Not super logical, but it is one way of doing that. $\endgroup$ – Paichu Nov 19 '15 at 4:52
  • $\begingroup$ You're right. I just fixed it. $\endgroup$ – Paichu Nov 19 '15 at 5:16
  • $\begingroup$ Also how can we assume $5a+b=-D$ ?? $\endgroup$ – Freelancer Nov 19 '15 at 5:17
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    $\begingroup$ $D\le 2$ actually means $D = -1$ or $D=-2$. Also, how does $a>0,b<0$ give the same thing as in (1)? $\endgroup$ – Joey Zou Nov 19 '15 at 7:50

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