positive roots. [Hint: Multiply by x + 1, which not change the number of positive roots.] 4. Show that the equation x if 5. Find the number of positive roots of the equation x³ + 2x³ – x² +x – 1 = 0. or an 6. From Descartes's rule of signs, conclude that the equation x2" –1 = 0 has 2n – 2 imaginary roots. 7. Without actually obtaining these roots, show that (a) x + 3x +7=0 and (b) x6 – 5x³ – 7x² + 8x + 20 = 0 lo iniog ou both possess imaginary roots. %3D %3D 8. Verify the following assertions. If all the coefficients of an equation are positive (a) and the equation involves no odd powers of x, then all its roots are imaginary. If all the coefficients of an equation are positive [h (b) and all terms involve odd powers of x, then zero is the only real root of the equation. Newton: The Principia Mathematica 8.3 For M.. of The Textbooks of Oughtred spe hea and Harriot wh the the will of the people. The smoldering antag into a conflict of the worst kind, the Great C lusibnsq 8 The Mechanical World:Descartes and Newton 8. (c) An equation with only positive coefficients che cannot have a positive root. noailuaed ro jo Oooiplaiabtylofle sidnin (a) The equation x³+a?x+b² = 0 has one negative and two imaginary roots if b0. (b) The equation x³ – a²x + b² = 0 has one negative root, while the other two roots are either imaginary or both positive. (c) The equation x4 + a²x² + b²x – c² = 0 has just two imaginary roots if c 0. corme 9. Prove that Modi be brur 10. Use Descartes's method to find the solutions of the following quartic equations: (a) x4 - 3x2 + 6x – 2 = 0. [Hint: The sextic Ok* + 17k2 – 36 = 0 has k? = 4 as solution 1 %3D
positive roots. [Hint: Multiply by x + 1, which not change the number of positive roots.] 4. Show that the equation x if 5. Find the number of positive roots of the equation x³ + 2x³ – x² +x – 1 = 0. or an 6. From Descartes's rule of signs, conclude that the equation x2" –1 = 0 has 2n – 2 imaginary roots. 7. Without actually obtaining these roots, show that (a) x + 3x +7=0 and (b) x6 – 5x³ – 7x² + 8x + 20 = 0 lo iniog ou both possess imaginary roots. %3D %3D 8. Verify the following assertions. If all the coefficients of an equation are positive (a) and the equation involves no odd powers of x, then all its roots are imaginary. If all the coefficients of an equation are positive [h (b) and all terms involve odd powers of x, then zero is the only real root of the equation. Newton: The Principia Mathematica 8.3 For M.. of The Textbooks of Oughtred spe hea and Harriot wh the the will of the people. The smoldering antag into a conflict of the worst kind, the Great C lusibnsq 8 The Mechanical World:Descartes and Newton 8. (c) An equation with only positive coefficients che cannot have a positive root. noailuaed ro jo Oooiplaiabtylofle sidnin (a) The equation x³+a?x+b² = 0 has one negative and two imaginary roots if b0. (b) The equation x³ – a²x + b² = 0 has one negative root, while the other two roots are either imaginary or both positive. (c) The equation x4 + a²x² + b²x – c² = 0 has just two imaginary roots if c 0. corme 9. Prove that Modi be brur 10. Use Descartes's method to find the solutions of the following quartic equations: (a) x4 - 3x2 + 6x – 2 = 0. [Hint: The sextic Ok* + 17k2 – 36 = 0 has k? = 4 as solution 1 %3D
Algebra & Trigonometry with Analytic Geometry
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Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 68E
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Topic Video
Question
Number 8b and c
![positive roots. [Hint: Multiply by x + 1, which
not change the number of positive roots.]
4. Show that the equation x
if
5. Find the number of positive roots of the equation
x³ + 2x³ – x² +x – 1 = 0.
or
an
6. From Descartes's rule of signs, conclude that the
equation x2" –1 = 0 has 2n – 2 imaginary roots.
7. Without actually obtaining these roots, show that
(a) x + 3x +7=0 and
(b) x6 – 5x³ – 7x² + 8x + 20 = 0
lo iniog ou
both possess imaginary roots.
%3D
%3D
8. Verify the following assertions.
If all the coefficients of an equation are positive
(a)
and the equation involves no odd powers of x,
then all its roots are imaginary.
If all the coefficients of an equation are positive
[h
(b)
and all terms involve odd powers of x, then zero
is the only real root of the equation.
Newton: The Principia Mathematica
8.3
For
M..
of
The Textbooks of Oughtred
spe
hea
and Harriot
wh
the
the will of the people. The smoldering antag
into a conflict of the worst kind, the Great C
lusibnsq](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc27b03fb-27ff-4129-a29b-052f3c3c0c2f%2Fa01b02ff-f0b1-48e1-aec7-39f06c9c5a13%2Fd62q11x.jpeg&w=3840&q=75)
Transcribed Image Text:positive roots. [Hint: Multiply by x + 1, which
not change the number of positive roots.]
4. Show that the equation x
if
5. Find the number of positive roots of the equation
x³ + 2x³ – x² +x – 1 = 0.
or
an
6. From Descartes's rule of signs, conclude that the
equation x2" –1 = 0 has 2n – 2 imaginary roots.
7. Without actually obtaining these roots, show that
(a) x + 3x +7=0 and
(b) x6 – 5x³ – 7x² + 8x + 20 = 0
lo iniog ou
both possess imaginary roots.
%3D
%3D
8. Verify the following assertions.
If all the coefficients of an equation are positive
(a)
and the equation involves no odd powers of x,
then all its roots are imaginary.
If all the coefficients of an equation are positive
[h
(b)
and all terms involve odd powers of x, then zero
is the only real root of the equation.
Newton: The Principia Mathematica
8.3
For
M..
of
The Textbooks of Oughtred
spe
hea
and Harriot
wh
the
the will of the people. The smoldering antag
into a conflict of the worst kind, the Great C
lusibnsq
Transcribed Image Text:8 The Mechanical World:Descartes and Newton
8.
(c)
An equation with only positive coefficients
che
cannot have a positive root.
noailuaed ro jo
Oooiplaiabtylofle
sidnin (a) The equation x³+a?x+b² = 0 has one
negative and two imaginary roots if b0.
(b) The equation x³ – a²x + b² = 0 has one
negative root, while the other two roots are either
imaginary or both positive.
(c) The equation x4 + a²x² + b²x – c² = 0 has just
two imaginary roots if c 0.
corme
9. Prove that
Modi be
brur
10. Use Descartes's method to find the solutions of the
following quartic equations:
(a) x4 - 3x2 + 6x – 2 = 0. [Hint: The sextic
Ok* + 17k2 – 36 = 0 has k? = 4 as
solution 1
%3D
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