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# Let x and y be the solutions to 7x^2+x-5=0. Compute (a-4)(b-4).

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Let j and k be the solutions to $$7x^2+x-5=0$$. Compute $$(j-4)(k-4)$$.

Apr 2, 2020
edited by mathmathj28  Apr 2, 2020

#1
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The solutions to 7x^2 + x - 5 = 0 are (-1 + i*sqrt(139))/14 and (-1 - i*sqrt(139))/14.  Then ((-1 + i*sqrt(139))/14 - 4)((-1 - i*sqrt(139))/14) - 4) = 111/7.

Apr 2, 2020
#2
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j and k are solutions means they  are the roots of the equations

Relation of roots to the polynomial (Quadratic equation in this case) is as follows:

j+k= $$-\frac{1}{7}$$ (1)

jk=$$-\frac{5}{7}$$ (2)

$$jk-4j-4k+16$$ (This is the expansion of (j-4)(k-4)

$$jk-4(j+k)+16$$  (Factored -4)

$$-\frac{5}{7}-4(-\frac{1}{7})+16=15.32$$ (Substitute (1), (2))

Apr 2, 2020
#3
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General rule:

For example: $$ax^2+bx+c=0$$

Sum of roots: $$-\frac{b}{a}$$

Product of roots: $$\frac{c}{a}$$

This stands for every polynomial and not restricted to a second degree only.

For instance,

$$ax^3+bx^2+cx+d=0$$ (Where r,s,q are the roots)

Sum of roots: $$-\frac{b}{a}$$

Sum of each pair of roots: $$rs+sq+rq=\frac{c}{a}$$

Product of roots=$$rsq=-\frac{d}{a}$$

Guest Apr 2, 2020
#4
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Oops, the calculation is 15.857 (not 15.32)

Guest Apr 2, 2020
#5
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Thanks so much!!

Apr 2, 2020