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The roots of $7x^2 + x - 5 = 0$ are $a$ and $b.$ Compute $(a - 4)(b - 4).$

 Jun 14, 2022
 #1
avatar+579 
+1

Using the quadratic formula gives:

 

\(_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:7\left(-5\right)}}{2\cdot \:7}\)

 

\(x_1=\frac{-1+\sqrt{141}}{2\cdot \:7},\:x_2=\frac{-1-\sqrt{141}}{2\cdot \:7}\)

 

\(x=\frac{-1+\sqrt{141}}{14},\:x=\frac{-1-\sqrt{141}}{14}\)

So if a and b are roots:

 

\((\frac{-1+\sqrt{141}}{14}-4)(\frac{-1-\sqrt{141}}{14}-4)\)

 

(EDIT:  Just you are calc if you don't know how to solve!!!)

 

Just solve that and you got the answer cool

 

-Vinculum

 Jun 14, 2022
edited by Vinculum  Jun 14, 2022
 #2
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0

by using vieta's formula, the answer is 80/7

 Jun 14, 2022
 #3
avatar+2448 
0

Expand: \((a-4)(b-4) = ab - 4a - 4b + 16 = ab - 4(a+b) + 16 \)

 

By Vieta's formula, we know that \(ab = -{5 \over 7 }\) and \(a+b = -{1 \over 7}\)

 

Now, plugging this in, we get: \(-{5 \over 7} -4(-{1 \over 7})+16 = {-{5 \over 7}} + { 4 \over 7}+16 = \color{brown}\boxed{15{ 6 \over 7 }}\)

 Jun 14, 2022
 #4
avatar+579 
+1

I completely forgot bout the Vieta's formula

 

I knew I did something wrong when I got those crazy numbers

 

 

Thanks, BuilderBoi laugh

 Jun 15, 2022
 #5
avatar+303 
+1

Why did you like your comment when you got it wrong?

hipie  Jun 16, 2022
 #6
avatar+579 
0

I did?

 Jun 16, 2022

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