19=h/3 -8 whats h??

Mathematics

1 answer:

creativ13 [48]3 years ago

8 0

First add the 8 to both sides, canceling out the 8 in the right and adding it to the 19 on the left, making it 27. Multiplying undoes division so, After that, multiply by three on both sides, canceling out the three on the right and multiply 27 by 3, making it 81.

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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

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