+2094 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.
+499 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
+499 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
