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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
+556
+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

-Vinculum

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

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

Jun 14, 2022
#3
+1829
+2

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
+556
+1

I completely forgot bout the Vieta's formula

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

Thanks, BuilderBoi

Jun 15, 2022
#5
+280
+1

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

hipie  Jun 16, 2022
#6
+556
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I did?

Jun 16, 2022