# solve applied problems by using quadratic equations?

A positive interger is one more than twice another. Their product is 78. Find the two integers. can you help me to solve this problem and explain how you did it. Thank you i need to understand how to do this word problems please give a hint of how can i solve them

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Let the 1st number be called by a variable 'x'.

Let the 2nd number be called by a variable 'y'

Then x =2y+1

and xy=78 or x = 78/y. Now substitute in the 1st equation.

You get 78/y = 2y+1

Multiply bith sides with 'y' . You get

78 = y(2y+1) = 2y^2+y

2y^2+y-78 =0

The solution for quadratic equation ax^2+bx+c =0 is

x =[-b+/-sqrt(b^2-4ac)]/2a

y = [-1 +/-sqrt(1-4*(2)*(-78))]/2*2

= [-1 +/-sqrt(1+624)]/4

=[-1+/-sqrt(625)]/4

=[-1+/-25]/4

=24/4 or -26/4

= 6 or -6.5

So x =2*6+1 =13 or

x = -2*6.5+1 = -12

• solve applied problems by using quadratic equations? this is the approach i expect from my students - not trial and error and to recognize that the equation can be factored

n

2n+1

n(2n+1) = 78

2n^2 +1n = 78

2n^2 +1n -78 = 0

now factor this : (2n+13)(n-6) = 0

then use the zero property on each bracket

2n+ 13 = 0 ==> n = -13/2 which is not a positive integer so i discard this

n-6 = 0 ==> n = 6

yep this is one solution and the other is

2n+1 = 2*6+1 = 13

Source(s): i'm a math teacher
• x= one integer

2x+1 = other integer

x(2x+1) = 78

2x^2 + x -78 = 0

x = (-1 +/- 25)/4 <-- used quadratic formula)

x = 24/4 = 6 = 1st number

2x+1 = 12+1 = 13 = 2nd number

• If a is the shorter leg and b is the longer leg... a²+b²=10² for the reason that b=a+2... a²+(a+2)² = one hundred a²+a²+4a+4 = one hundred 2a²+4a+4=one hundred 2a²+4a-ninety six=0 2(a²+2a-40 8) = 0 2(a-6)(a+8) = 0 a=6, a=-8 for the reason that we are attempting to discover the leg of a triangle, a=-8 would not make experience. So a=6. the respond is B.