Question

For what value of c does x^2−2x−c=4 have exactly one real solution? PLEASE HELP!!!!!!!

Answer 1

Answer:

-5

Step-by-step explanation:

Moving all terms of the quadratic to one side, we have

$x^2-2x-(c+4)=0$.

A quadratic has one real solution when the discriminant is equal to 0. In a quadratic $ax^2+bx+d$, the discriminant is $\sqrt{b^2-4ad}$.

(The discriminant is more commonly known as $\sqrt{b^2-4ac}$, but I changed the variable since we already have a $c$ in the quadratic given.)

In the quadratic above, we have $a=1$, $b=-2$, and $d=-(c+4)$. Plugging this into the formula for the discriminant, we have

$\sqrt{(-2)^2-4(1)(-(c+4))$.

Using the distributive property to expand and simplifying, the expression becomes

$\sqrt{4-4(-c-4)}=\sqrt{4+4c+16}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{20+4c}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{4}\cdot\sqrt{5+c}\\~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{c+5}.$

Setting the discriminant equal to 0 gives

$2\sqrt{c+5}=0$.

We can then solve the equation as usual: first, divide by 2 on both sides:

$\sqrt{c+5}=0$.

Squaring both sides gives

$c+5=0$,

and subtracting 5 from both sides, we have

$\boxed{c=-5}.$