Answer:
-5
Step-by-step explanation:
Moving all terms of the quadratic to one side, we have
[tex]x^2-2x-(c+4)=0[/tex].
A quadratic has one real solution when the discriminant is equal to 0. In a quadratic [tex]ax^2+bx+d[/tex], the discriminant is [tex]\sqrt{b^2-4ad}[/tex].
(The discriminant is more commonly known as [tex]\sqrt{b^2-4ac}[/tex], but I changed the variable since we already have a [tex]c[/tex] in the quadratic given.)
In the quadratic above, we have [tex]a=1[/tex], [tex]b=-2[/tex], and [tex]d=-(c+4)[/tex]. Plugging this into the formula for the discriminant, we have
[tex]\sqrt{(-2)^2-4(1)(-(c+4))[/tex].
Using the distributive property to expand and simplifying, the expression becomes
[tex]\sqrt{4-4(-c-4)}=\sqrt{4+4c+16}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{20+4c}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{4}\cdot\sqrt{5+c}\\~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{c+5}.[/tex]
Setting the discriminant equal to 0 gives
[tex]2\sqrt{c+5}=0[/tex].
We can then solve the equation as usual: first, divide by 2 on both sides:
[tex]\sqrt{c+5}=0[/tex].
Squaring both sides gives
[tex]c+5=0[/tex],
and subtracting 5 from both sides, we have
[tex]\boxed{c=-5}.[/tex]
