SOLUTION: factor completely (x+3)^2-10(x+3)+24 write a quadratic equation and answer the following: the shortest leg of a right triangle is two less than the hypotenuse. the hypotenuse

Question 324557: factor completely (x+3)^2-10(x+3)+24

write a quadratic equation and answer the following: the shortest leg of a right triangle is two less than the hypotenuse. the hypotenuse is one longer than the longest leg.
thank you
Found 2 solutions by katealdridge, mathslover:
Answer by katealdridge(100) (Show Source): You can put this solution on YOUR website!
1. Factor completely
suppose (x+3)=y instead
Factor-look for 2 numbers that mult to 24 and add or subtract to get -10. They are -6 and -4.

Now substitute (x+3) back in for y.
simplify

2. Write a quadratic equation and answer the following: the shortest leg of a right triangle is two less than the hypotenuse. The hypotenuse is one longer than the longest leg.
let shortest leg=a, let longer leg=b, let hypotenuse=c

substitute b+1 for c in first equation

simplified
to summarize: ,,a, b, and c are all in terms of b.
Pythagorean theorem
Pyth thm with b-1 substituted for a and b+1 substituted for c.
FOIL and
simplified
b^2 subtracted from both sides
2b and 1 subtracted from both sides
factor out b
set each factor equal to zero and solve
Since this is a real life geometry problem, b=0 does not make sense and is therefore excluded as an answer.
and 4 substituted in for b in a=b-1 and c=b+1 equations
Answer by mathslover(157) (Show Source): You can put this solution on YOUR website!
(x+3)^2-10(x+3)+24
Let's replace x+3 by A
So, the given expression would transform to
A^2 -10A + 24
Rewriting it as
A^2 - 6A -4A + 24
A(A-6) -4(A-6)
Taking A-6 as the common factor,
(A-6)(A-4)
Substituting A=x+3 back into the expression we get
(x+3-6)(x+3-4)
(x-3)(x-1) is the required factor.
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write a quadratic equation and answer the following: the shortest leg of a right triangle is two less than the hypotenuse. the hypotenuse is one longer than the longest leg.
Let the longest leg be x
So the hypotenuse is (x+1) {the hypotenuse is one longer than the longest leg}
And the shortest leg is (x+1)-2=x-1 {the shortest leg of a right triangle is two
less than the hypotenuse}
Using Pythagoras's theorem,
hypotenuse^2 = shorter leg^2 + longer leg^2
we get
(x+1)^2 = (x-1)^2 + x^2
Expanding the LHS and the RHS
x^2 + 2x+ 1 = x^2 -2x+1 + x^2
Moving the Right hand side expression to Left hand side we get
x^2 + 2x+ 1 - (x^2 -2x+1 + x^2)= 0
grouping the like terms,
(x^2 - x^2 -x^2) + (2x + 2x ) + (1 -1)=0
-x^2 + 4x =0
-x(x-4)=0
which gives x=0 and x=4 as two possible solutions. But we cannot have a triangle with side of length 0. So discarding x=0 we get the other solution as x=4
Hence, the sides of the triangle are
The longest side=4
The shorter side=4-1=3
hypotenuse=4+1=5

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