Question

Find all positive integer values of $c$ such that the equation $x^2-7x+c=0$ only has roots that are real and rational. Express them in decreasing order, separated by commas.

Answer 1

Answer:

12, 10, 6 (descending order)

Step-by-step explanation:

$ax^2+bx+c=0$ has real and rational roots if $b^2-4ac$ is a nonnegative perfect square. $b^2-4ac$ is called the discriminant.

Let's calculate the discriminant of $x^2-7x+c=0$.

$b^2-4ac$

$(-7)^2-4(1)(c)$

$49-4c$

We want $49-4c$ to be be positive or a zero perfect square.

So let's first solve the inequality:

$49-4c \ge 0$

Add $4c$ on both sides:

$49 \ge 4c$

$4c \le 49$

Divide both sides be 4:

$c \le \frac{49}{4}$

$c \le 12.25$.

We know $c$ has to be a positive integer.

So we only need to evaluate $49-4c$ for 12 numbers.

$c=1$ returns $49-4(1)=49-4=45$ which is not a perfect square.

$c=2$ returns $49-4(2)=49-8=41$ which is not a perfect square.

$c=3$ returns $49-4(3)=49-12=37$ which is not a perfect square.

$c=4$ returns $49-4(4)=49-16=33$ which is not a perfect square.

$c=5$ returns $49-4(5)=49-20=29$ which is not a perfect square.

$c=6$ returns $49-4(6)=49-24=25$ which is a perfect square.

$c=7$ returns $49-4(7)=49-28=21$ which is not a perfect square.

$c=8$ returns $49-4(8)=49-32=17$ which is not a perfect square.

$c=9$ returns $49-4(9)=49-36=13$ which is not a perfect square.

$c=10$ returns $49-4(10)=49-40=9$ which is a perfect square.

$c=11$ returns $49-4(11)=49-44=5$ which is not a perfect square.

$c=12$ returns $49-4(12)=49-48=1$ which is a perfect square.