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The quadratic x^2+3/2x-1 has the following unexpected property: the roots, which are 1/2 and -2, are one less than the final two coefficients. Now find a quadratic with leading term x^2 such that the final two coefficients are both non-zero, and the roots are one more than these coefficients.

 

Any help will be greatly appreciated! Thanks in advance.

 Mar 23, 2020

Best Answer 

 #1
avatar+497 
+2

Name the quadratic you try to find as :

 

\(x^2+ax+b\)

Next we define the roots as:

x1 and x2

 

By vietas formulas:

 

x1 + x2 = -a

x1 * x2 = b

Since we know that the roots are one more than the last two coefficients respectively, we have:

x1 = a + 1

x2 = b+ 1

Substituting back in, we get:

a + 1 + (b+1) = -a

a+b + 2 = -a

2a = -b-2

a = -b/2 - 1

 

(a+1) * (b+1) = b

(-b/2 -1 + 1) * (b+1) = b

(-b/2) * (b+1) = b

(-b2/2 - b/2) = b

b2/2 + b/2 = -b

Dividing by b on both sides, we get:

b/2 + 1/2 = -1

b/2 = -3/2

b = -3

plugging back into our first equation, we get:

(a + 1) + (-3 + 1) = -a

a + 1 -2 = -a

a -1 = -a

2a = 1

a = (1/2)

our two roots are 1/2 and -3

 Mar 23, 2020
 #1
avatar+497 
+2
Best Answer

Name the quadratic you try to find as :

 

\(x^2+ax+b\)

Next we define the roots as:

x1 and x2

 

By vietas formulas:

 

x1 + x2 = -a

x1 * x2 = b

Since we know that the roots are one more than the last two coefficients respectively, we have:

x1 = a + 1

x2 = b+ 1

Substituting back in, we get:

a + 1 + (b+1) = -a

a+b + 2 = -a

2a = -b-2

a = -b/2 - 1

 

(a+1) * (b+1) = b

(-b/2 -1 + 1) * (b+1) = b

(-b/2) * (b+1) = b

(-b2/2 - b/2) = b

b2/2 + b/2 = -b

Dividing by b on both sides, we get:

b/2 + 1/2 = -1

b/2 = -3/2

b = -3

plugging back into our first equation, we get:

(a + 1) + (-3 + 1) = -a

a + 1 -2 = -a

a -1 = -a

2a = 1

a = (1/2)

our two roots are 1/2 and -3

jfan17 Mar 23, 2020
 #2
avatar+2314 
+1

Thanks so much, jfan! That's impressive!

 Mar 23, 2020
 #3
avatar+111456 
+1

I agree.....very nice, jfan  !!!!!

 

 

cool cool cool

CPhill  Mar 23, 2020

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