Quadratic with a Repeated Root

In order to have a repeated root, $b^2-4ac$ must be equals to $0$ . Therefore, $\frac{b^2}{4}=ac$ and $b^2$ must be divisible by $4$ . Therefore, $b$ must be even.

When $b = 10$ , $ac=\frac{10^2}{4}=25$ . So the only possible permutation is $(5, 10, 5)$ .

When $b = 8$ , $ac=\frac{8^2}{4}=16$ . Therefore, the ordered pair $(a,c)$ can be: $(4,4)$ , $(2,8)$ and $(8,2)$ . There are thus 3 possible ordered triplets.

When $b = 6$ , $ac=\frac{6^2}{4}=9$ . Therefore, the ordered pair $(a,c)$ can be: $(3,3)$ , $(1,9)$ and $(9,1)$ . There are thus 3 possible ordered triplets.

When $b = 4$ , $ac=\frac{4^2}{4}=4$ . Therefore, the ordered pair $(a,c)$ can be: $(2,2)$ , $(1,4)$ , and $(4,1)$ . There are thus 3 possible ordered triplets.

When $b = 2$ , $ac=\frac{2^2}{4}=1$ . So the only possible permutation is $(1, 2, 1)$ .

Therefore, there are, in total, $1+3+3+3+1=11$ ordered triples.

For the quadratic to have a repeated root, the discriminant must equal $0,$ that is, we must have $b^2-4ac = 0,$ or

$b^2=4ac.$

From this we see that $b$ must be even, and we take cases:

• $b = 2$ gives $ac = 1,$ so $a=c=1$ for one possibility.

• $b=4$ gives $ac = 4,$ so $(a,c) = (1,4), (2,2), (4,1),$ for three possibilities.

• $b = 6$ gives $ac = 9,$ so $(a,c) = (1,9), (3,3), (9,1),$ for three more possibilities.

• $b = 8$ gives $ac = 16,$ and since $a, c\le 10,$ the only pairs are $(a,c) = (2,8), (4, 4), (8, 2),$ for three more possibilities.

• $b = 10$ gives $ac = 25,$ so the only pair is $a=c=5.$

In total, there are $1+3+3+3+1 = \boxed{11}$ ordered triples.

If equation has repeating, then it is said to have equal roots.

Equal Roots in turn means - Discriminant = 0.

$\ b^{2} - 4ac$ = 0

Implies,

$\ b^{2}$ = 4ac

Minimum value of b is 2,

which implies,

$\ 4 = 4ac$

This gives us one ordered triple = (1 , 2 , 1)

Then another condition arises where $\ b$ is even only.

Now b = 4,

4 = $\ ac$

Cases are $\ a$ = 1 , 2 , 4 and $\ c$ = 4 , 2 , 1 respectively .

Now we get 3 more triples, they being: (1 , 4 , 4) , (2 , 4 , 2) , (4 , 4, 1)

Now $\ b$ = 6,

$\ 9 = ac$

Now we get 3 more, $\ a$ = 1 , 3 , 9 and $\ c$ = 9 , 3 , 1 respectively

The triples are: (1 , 6 , 9) , (3 , 6 , 3) , (9 , 6 , 1)

Now $\ b$ = 8,

Three more: (2 , 8 , 8) , (4 , 8 , 4) , (8 , 8 , 2)

Now $\ b$ = 10 (max)

One Triple arises:

(5 , 10 , 5)

Totally $\ 1 + 3 + 3 + 3 + 1$ = $\ 11$

If it should have a repeated root then b^2-4 a c=0 ->b^2=4ac ->b=2 sqrt(a c). sqrt(a c) is an integer only when a c=1,4,9,16,25(above 25 you will get b>10).So 1)If a c=1,(1,2,1) 2)If a c=4 (2,4,2),(4,4,1),(1,4,4) 3)If a c=9 (9,6,1),(1,6,9),(3,6,3) 4)If a c=16 (4,8,4),(2,8,8),(8,8,2) 5)If a*c=25 (5,10,5) So total number of solutions=11

We are given $b^2 = 4ac$ , clearly $2 \mid b$ , now consider five cases to get following triplets $(a,b,c)$ $b = 2 \Rightarrow ac = 1: (1,2,1) \\ b = 4 \Rightarrow ac = 4: (1,4,4), (4,4,1), (2,4,2) \\ b = 6 \Rightarrow ac = 9: (1,6,9), (9,6,1), (3,6,3) \\ b = 8 \Rightarrow ac = 16: (2,8,8), (4,8,4), (8,8,2) \\ b = 10 \Rightarrow ac = 25: (5,10,5)$

Hence the number of triplets is $\boxed{11}$

$b^{2}-4ac=0$ hence the solution vector will be $(a b c)= a (2 \sqrt{ac}) c$ so the there will be 11 solution of the case like , a=1, c=2 and 9, a=2, c=2 and 8, a=3, c=3, a=4, c=4 and 9, a=5, c=5, a=6, c=6, a=7 c= nothing no solution here, a=8 c=2, a=9 c=1m, a=10 no solutio n hence total 11 solution

1+2+2+1+1+2+1+1=11

For a root to be repeated, the determinant $b^{2} - 4ac$ has to be zero. This means that $4ac$ , with a and c integers, is a square. Since no integers allow $4ac$ to be equal to the square of an odd integer, I counted a and c values that gave the square of any even integer up to 10. There was one $(a,b,c)$ triplet for $2^{2} = 4$ , one triplet for $10^{2} = 100$ , and three triplets each for the squares of the remaining even integers between one and ten. This added up to 11 triplets.

For a quadratic to have a repeated root, the discriminant $b^2 - 4ac$ must equal 0. This tells us that $b = 2 \sqrt{ac}$ . We consider the possible values for $b$ and look at all possible integer solutions for $a$ and $c$ . If $b = 2$ then $\sqrt{ac} = 1$ , so $(a,c) = (1,1)$ . If $b = 4$ , then $\sqrt{ac} = 2$ and so $(a,c) = (1,4), (2,2), \mbox{ or } (4,1)$ . If $b = 6$ then $\sqrt{ac} = 3$ and so $(a,c) = (1,9), (3,3), \mbox{ or } (9,1)$ . If $b = 8$ then $\sqrt{ac} = 4$ and so $(a,c) = (1,16), (2,8), (4,4), (8,2), \mbox{ or } (16,1)$ . If $b = 10$ then $\sqrt{ac} = 5$ and so $(a,c) = (1,25), (5,5), \mbox{ or } (25,1)$ . This gives a total of $1 + 3 + 3 + 5 + 3 = 15$ possible solutions. 4 of these solutions have values of $a$ or $c$ that is greater than 10, so these are not valid solutions. Thus, there are 11 possible triples $(a,b,c)$ .

Since b^2 - 4 a c = 0, and they are all positive integers 1 t0 10, inclusive.
b = 2 sqrt(a c) . Thus max a*c can be 25 only. Hence we take a=c with values 1 to 5....Five.
If a = 1, we can have c = 4 and c = 9. So also c = 1, a = 4 and 9........................................Four
If a = 2, we can have c = 8. So also c = 2, a = 8 ............................. .............. .......Two
a=3 , must have c = 3 we have done this, so also for a= 4, 5.
a = 6, 7, 10 have no matching c. a = 8, 9 has been covered. ........................................... 11 ........

Function has repeated roots when it's discriminant is 0 so: $b^2-4ab=0 \rightarrow b^2=4ab$ we notice that $b^2$ must be divisible by $4$ and that is when $b=\{ 2,4,6,8,10\}$

1. $b=2 \\ 4=4ac \\ ac=1 \\ a=1 \wedge c=1$

2. $b=4 \\ 16 =4ac \\ ac=4 \\ (a_1=2 \wedge c_1 =2) \lor (a_2=4 \wedge c_2=1) \lor (a_3=1 \wedge c_3=4)$

3. $b=6 \\ 36 =4ac \\ ac=9 \\ (a_1=3 \wedge c_1 = 3) \lor (a_2=9 \wedge c_2=1)\lor (a_3=1 \wedge c_3=9)$

4. $b=8 \\ 64 = 4ac \\ ac=16 \\ (a_1=4 \wedge c_1=4) \lor (a_2=8 \wedge c_2=2)\lor (a_3=2 \wedge c_3=8)$

5. $b=10 \\ 100=4ac \\ ac=25 \\ a=5 \wedge b=5$

if we add all the solutions we got we get: $1+3+3+3+1=\boxed{11}$

for values where the roots are equal the discriminant is 0. using discriminant, b^2-4ac, given it's = to 0 b^2=4ac now consider all values of b 1 to 10 only even numbers should be used as b cant be odd as odd numbers arent divisible by 4 2^2=4ac a and c only have one possible order 1,1 4^2=4ac a and c have 3 possible orders 4,1 1,4 2,2 6^2=4ac a and c have 3 values for this value of b 3,3 9,1 1,9 8^2=4ac a and c have 3 orders of a and c 4,4 8,2 2,8 10^2=4ac a and c only have one order 5,5 notice that 25,1 and 1,25 cant occur as a,b,c <11 and integer

1+3+3+3+1=11

11 solutions

For repeated roots, b^2 = 4ac. Since 4 | b^2, b can take values 2, 4, 6, 8 or 10. When b = 2, we have ac = 1. Therefore (a,b,c) = (1,2,1) When b = 4, we have ac = 4. Therefore (a,b,c) = (1,4,4) or (4,4,1) or (2,4,2) When b = 6, we have ac = 9. Therefore (a,b,c) = (1,6,9) or (9,6,1) or (3,6,3) When b = 8, we have ac = 16. Therefore (a,b,c) = (2,8,8) or (8,8,2) or (4,8,4) When b = 10, we have ac = 25. Therefore (a,b,c = (5,10,5)

We have exhausted all possible lists and therefore there are 11 such ordered triples.

By the formula for the squares of quadratic roots, since both roots are equal: (-b+sqrt(b^2-4ac))/2a = (-b-sqrt(b^2-4ac))/2a. Therefore, b^2=4ac, so b=2sqrt(ac). So, since b is an integer, we only consider the cases when, choosing two numbers from 1 to 10 (ac), their product is a perfect square. We see that 11 numbers satisfy this, hence the answer is 11.

Condition for repeated roots: $b^2 = 4ac$ There are total of 11 ordered pairs in the range $[1,10]$ satisfying this condition