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3 years ago

Solve the quadratic equation using the quadratic formula: 3x^2 - 6x = 5

Mathematics

1 answer:

Fantom [35]3 years ago

8 0

Answer:

$x_{1} = 1+\frac{2\sqrt{6}}{3} \\\\x_{2} = 1-\frac{2\sqrt{6}}{3}$

Step-by-step explanation:

3x^2 - 6x - 5=0

quadratic formula:

$x=\frac{-b+-\sqrt{b^{2}-4*a*c}}{2*a}$

a=3 b=-6 c=-5

$x=\frac{-(-6)+-\sqrt{(-6)^{2}-4*3*-5}}{2*3}$

$x=\frac{6+-\sqrt{36+60}}{6} \\\\x=\frac{6+-\sqrt{96}}{6} \\\\x=\frac{6+-\sqrt{2^{2}*2^{2}*2*3 }}{6} \\\\x=\frac{6+-4\sqrt{6}}{6} \\$

so we have:

$x_{1} = 1+\frac{2\sqrt{6}}{3} \\\\x_{2} = 1-\frac{2\sqrt{6}}{3}$

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grandymaker [24]

The answer is: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
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Explanation:
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Given the original function:

" c(y) = (5/9) (x − 32) " ; in which "x = f" ; and "y = c(f) " ;
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→ Write the original function as: " y = (5/9) (x − 32) " ;

Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
________________________________________________________
x = (5/9) (y − 32) ;

Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ;
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→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
"left-hand side" of the equation:

We have:

→ x = " ( $\frac{5}{9}$ ) * (y − 32) " ;

Let us simplify the "right-hand side" of the equation:
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Note the "distributive property" of multiplication:
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a(b + c) = ab + ac ; AND:

a(b – c) = ab – ac .
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As such:
__________________________________________

" ( $\frac{5}{9}$ ) * (y − 32) " ;

= [ ( $\frac{5}{9}$ ) * y ] − [ ( $\frac{5}{9}$ ) * (32) ] ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{5}{9}$ ) * ( $\frac{32}{1})$ " ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{(5*32)}{(9*1)}$ ] ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{(160)}{(9)}$ ] ;

= [ ( $\frac{5y}{9}$ ) ] − [ ( $\frac{(160)}{(9)}$ ] ;

= [ $\frac{(5y-160)}{9}$ ] ;
_______________________________________________
And rewrite as:

→ " x = $\frac{(5y-160)}{9}$ " ;

We want to rewrite this; solving for "y"; with "y" isolated as a "single variable" on the "left-hand side" of the equation ;

We have:

→ " x = $\frac{(5y-160)}{9}$ " ;

↔ " $\frac{(5y-160)}{9}$ = x ;

Multiply both sides of the equation by "9" ;

9 * $\frac{(5y-160)}{9}$ = x * 9 ;

to get:

→ 5y − 160 = 9x ;

Now, add "160" to each side of the equation; as follows:
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→ 5y − 160 + 160 = 9x + 160 ;

to get:

→ 5y = 9x + 160 ;

Now, divided Each side of the equation by "5" ;
to isolate "y" on one side of the equation; & to solve for "y" ;

→ 5y / 5 = (9y + 160) / 5 ;

to get:

→ y = (9/5)x + (160/5) ;

→ y = (9/5)x + 32 ;

→ Now, remember we had substituted: "y" for "c(f)" ;

Now that we have the "equation for the inverse" ;
→ which is: " (9/5)x + 32" ;

Remember that for the original ("non-inverse" equation); "y" was used in place of "c(f)" . We have the "inverse equation"; so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as: "f(c)" .

Note that "x = c" ;
_____________________________________________________
So, the inverse function is: " f(c) = (9/5) c + 32 " .
_____________________________________________________

The answer is: " f(c) = $\frac{9}{5}$ c + 32 " ;
_____________________________________________________
→ which is:

→ Answer choice: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
_____________________________________________________

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