+0

# Help!!

0
38
2

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

Thank you! (it's not 80/7)

Jun 3, 2022

#1
0

By the quadratic formula, a and b are $$\frac{-1 \pm \sqrt{141}}{4}$$

Plug these into (a - 4)(b - 4), and you get 103/7

Jun 3, 2022
#2
0

There are two methods to solve this.

First method:

Expand:

$$(a-4)(b-4)=ab-4a-4b+16=ab-4(a+b)+16$$

Then, if a and b are the roots of $$7x^2+x-5=0$$, then by Vietas' formulae:

$$a+b=-\frac{1}{7}$$  and $$ab=-\frac{5}{7}$$

Therefore:

$$(a-4)(b-4)=-\frac{5}{7}-4(-\frac{1}{7})+16=\frac{111}{7}$$ is the desired answer.

Second method:

If a and b are the roots of $$7x^2+x-5=0$$

Then, we do the following substitution: let $$y=x-4$$, hence $$x=y+4$$

So:

$$7(y+4)^2+(y+4)-5=0 \iff 7(y^2+8y+16)+y+4-5=7y^2+56y+112+y-1=7y^2+57y+111$$

Therefore, the product of the roots (Which is now (a-4)(b-4), as we transformed the quadratic with the roots a and b to another quadratic with roots a-4 and b-4.) is again, by Vietas' formulae: $$\frac{111}{7}$$.

Jun 4, 2022