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TopNewestI think, that by Descartes' rule of signs, the signs of the coefficients must interleave for the solutions to be positive, so for the quadratic equation \(a\) and \(c\) must have equal signs and \(b\) must have opposite sign; similarity for the cubic equation, \(a\) and \(c\) must have equal signs, and \(b\) and \(d\) must have equal signs but opposite from \(a\) and \(c\).

Now, the second condition is that the discriminant must be

greater than 0, that is because the discriminant is just the product of the squaresof every possible difference of two roots at time, all that multiplied by \(a^{2n-2}\), where \(a\) is the leading coefficient. For example, for a second degree equation: \(D=a^2 (x_1-x_2)^2\), and for a third degree equation: \(D=a^4 (x_1-x_2)^2 (x_1-x_3)^2 (x_2-x_3)^2\). It's clearly that if at least one root is repeated, the whole discriminant will be 0. And if there are a pair of complex roots it will be negative. But, these discriminants now in terms of the coefficients are: \(D=b^2-4ac\) and \(D=b^2 c^2 - 4ac^3 - 4b^3 d + 18abcd - 27a^2 d^2 \).So, in both cases, \(D>0\).I think that in these grades these are the sufficient and necesary conditions for all roots to be distinct and positive, but in higher degrees I'm not sure. The proofs of how to obtain the discrimimants are a little hard and I will not give them, also I'm writing this from my phone, and it's a pain, so if there are any mistakes, I'm sorry. – Alan Enrique Ontiveros Salazar · 3 years, 3 months ago

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Firstly, in the quadratic and cubic case, a positive discriminant is a necessary and sufficient condition that the roots are real.

Secondly, Descartes' rule of signs tells us that if the signs are alternating, then there are (at most) 0 negative roots.

Hence, all the roots are positive and real!

The converse is obvious. If the roots are positive reals, then the discriminant is positive, and the signs alternate. – Calvin Lin Staff · 3 years, 3 months ago

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For the quadratic equation, 2 conditions arise-

If \(a>0\), then \(c>0\) and \(b^{2} - 4ac>0\). Also, \( \dfrac{-b}{2a}>0\) which means that \(b<0\)

If \(a<0\), then \(c<0\) and \(b^{2} - 4ac>0\). Also, \( \dfrac{-b}{2a}>0\) which means that \(b>0\)

For the cubic equation, 2 conditions arise-

Let \(f(x) = ax^{3} + bx^{2} + cx + d\)

If the cubic has 3 roots, then \(f'(x)\) has \(\textbf{2 positive real roots}\). Let they be \(p,q\). The conditions for it can be found from above.

For three roots,

\(f(p) . f(q) <0\)

Assuming that the condition is satisfied, then 2 cases arise-

If \(a<0\), then \(d>0\)

If \(a>0\), then \(d<0\) – Avineil Jain · 3 years, 3 months ago

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– Calvin Lin Staff · 3 years, 3 months ago

For the quadratic case, can you explain why those are necessary and sufficient conditions? What would be the best way of interpreting it?Log in to reply

Product of roots is \(\dfrac{c}{a}\) which means \(a\) and \(c\) are of the same sign.

Sum of roots is \(\dfrac{-b}{a}\) which means \(a\) and \(b\) are of opposite signs.

Also, the discriminant \(b^{2} - 4ac\) should be positive for two real roots. – Avineil Jain · 3 years, 3 months ago

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Even if \(f'(x)\) has 2 roots, it doesn't guarantee that \(f(x)\) has two more real roots.

For example: look at this one. – Pranav Arora · 3 years, 3 months ago

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– Avineil Jain · 3 years, 3 months ago

All right, i will modify it.Log in to reply

if x and y are real numbers satisfying x^3-3x^2+5x-17=0 and y^3-3y^2+5y+11=0 then the value of x+y=

– Archana Desai · 2 years, 1 month ago______Log in to reply

This is with regard to second part, although the approach could also be used for quadratic case. The general cubic equation is

\(a{ x }^{ 3 }+b{ x }^{ 2 }+cx+d=0\)

Convert the general cubic equation into depressed cubic.Thus

\(t^{3}-mt-c=0\)

with \(t=x+b/(3a)\)

Then we may consider this to be the intersection of two curves namely \(y=t^{3}\)and a straight line \(y=mt+c\). Now plotting these two curves one clearly sees that for 3 real roots the slope m of the line must be positive.Thus condition one is

\(m>0\)

This is equivalent to the condition that the derivative of the cubic polynomial must have two roots.Next if there were two repeated roots then the slope of the line and tangent are equal. Equating both we get

\({ t }_{ 0 }=\pm \sqrt { \frac { m }{ 3 } }\) .

Substituting t0 back into the straight line equation we get a value for the y-intercept for this special case and noting that the intercept \(\left| c \right| \) must be less than this for 3 unequal real roots to exist.Thus we have the condition two

\(\left| c \right| \le \quad \frac { 2m }{ 3 } \sqrt { \frac { m }{ 3 } } \)

This is equivalent to the discriminant condition of the cubic equation. Next we turn to the question of positive roots. This is equivalent to

\( t>\frac { b }{ 3a } \)

Now we note that if the depressed cubic has 3 real roots then by necessity it will have one negative root (since sum of roots is zero in depressed cubic & by same logic it will also atleast have one positive root). So for positive real roots of the original cubic we need to have

\(\frac { b }{ a } <0\)

This third condition is only necessary not sufficient.Next we ensure that the smallest root of depressed cubic is greater than b/3a . For this we suppose a straight line of slope m passes through the point \(\left( \frac { b }{ 3a } ,{ \left( \frac { b }{ 3a } \right) }^{ 3 } \right)\) then the y intercept of this line comes out to be

\({ c }_{ 0 }={ \left( \frac { b }{ 3a } \right) }^{ 3 }-m\frac { b }{ 3a } \)

This c0 must be lesser than y intercept of the straight line for roots of depressed cubic to be greater than b/3a.Thus condition four is

\({ c }_{ 0 }<c\).

Once all the four conditions are satisfied simultaneously we will have positive unequal roots for the original cubic. – Sushant Vijayan · 3 years, 3 months ago

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– Sushant Vijayan · 3 years, 3 months ago

Condition 4 when simplified leads to d/a<0.Thus condition three and four are equivalent to descartes rule of signs.Log in to reply

– Sumit Das · 3 years, 3 months ago

Shusant I think u made out a spledid explanation thanks for the ri8 solution.Log in to reply

Cubic Equation x³+ax²+bx+c=0 with real coefficients a,b,c have ONLY REAL ROOTS iff

1) (2a³-9ab+27c)² ≤ 4(a²-3b)³

2) a² ≥ 3b

"+" to be 3 POSITIVE must be by Vietta

3) a<0

4) b>0

5) c<0 – Sladjan Stankovik · 1 year, 9 months ago

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b sqred - 4ac>0 – Prashant Patel · 3 years, 3 months ago

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guys, I hate to rain on your parade but can you elaborate this much clearer and simpler for me because I can't understand a word you're typing. please??? – Jay Cyril Mijares · 3 years, 3 months ago

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This makes me think about the problems I posted about Brilli the Ant playing the games against Brian Till. – Sharky Kesa · 3 years, 3 months ago

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– Calvin Lin Staff · 3 years, 3 months ago

Yes, that allows us to say that there are 2 real roots. But must the roots be positive? How can we ensure that these roots are positive?Log in to reply

– Bhargav Varshney · 3 years, 3 months ago

Yes, you are right. I will modify this.Log in to reply

a and c having same sign b is different – Hany Ganzy · 3 years, 3 months ago

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– Calvin Lin Staff · 3 years, 3 months ago

I agree that is necessary, but it is not sufficient. For example \( x^2 - x + 1 = 0 \) does not have 2 positive real roots.Log in to reply

What is the meaning of all these things i.e possible real roots and all ,these thing are not taught in my school – Aman Real · 3 years, 3 months ago

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Real numbers are just the set of rational and irrational numbers. – Sharky Kesa · 3 years, 3 months ago

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