If f(x) = 2x - 5 and g(x) = x + 52, then f(g(x)) can be deduced by placing g(x) in the spot of x in the f(x) equation as follows:

f(g(x)) = 2(g(x)) - 5

Since we know g(x) = x + 52, let's plug it in:

f(g(x)) = 2(x + 52) - 5
f(g(x)) = 2x + 104 - 5
f(g(x)) = 2x + 99

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2 years ago

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

Paul [167]

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

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