Mathematics Theorems Archives – Mathemafia https://mathemafia.com/category/mathematics-theorems/ Mathematics is like music - Enjoy it!!! Tue, 25 Mar 2025 09:15:29 +0000 en-US hourly 1 https://wordpress.org/?v=7.0.2 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Mathematics Theorems Archives – Mathemafia https://mathemafia.com/category/mathematics-theorems/ 32 32 SSA is not a valid congruence criterion https://mathemafia.com/ssa-is-not-a-valid-congruence-criterion/ Sun, 09 Mar 2025 12:17:43 +0000 https://mathemafia.com/?p=13890 A witty explanation and proof by mathemafia that SSA is not a valid congruence criterion when it comes to the congruence of triangles in geometry.

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SSA is not a valid congruence criterion

Student: What is SSA?

Mathemafia – I call it Side Side Aaaaaaaaaa…..hhhhhhΒ !!!

Not funny, huh!! But effective 🎯 enough to remember 🧠.

If you want to read, it may take 5 minutes. If you want to save time, watch the short video and then send me half πŸŒ“ of the money you save.Β 

You know… ⏰ Time is πŸ’° Money πŸ˜‚ 🀣 πŸ˜†

If you still insist, written below is the explanation why SSA is not a valid congruence criterion. 🀿 πŸ‘€ ❀️ πŸ’ƒ

But First, Let’s Sing it!!!

As I always say – Mathematics is like Music. Enjoy it!!! But don’t enjoy this one alone πŸ₯³. This piece of music (SSA) is a party πŸŽ‰ song 🎢. So enjoy it with your friends. πŸ’ƒπŸ•ΊπŸ’ƒπŸ•Ί. (I can’t ask in more creative 🎭 way to share πŸ”— it. Hmmmmm… Let me say it clearly then… I BEG YOU. SHARE it pls. 😭 πŸ™ πŸ˜­πŸ™ )Β 

SSA – Sing Sing Aaaaaaaa…..Β 

SSA – Share Share Aaaaaaaa…..Β 

SSA – Subscribe Subscribe Aaaaaaaa…..Β 

StudentΒ – Sir, how about if we reverse the abbreviation 🀭 πŸ™ˆ πŸ˜† πŸ‘ 🫏

Mathemafia – Wow!!! That’s much more creative. 🫏 🫏 

Explanation

Let us draw an obtuse angled triangle, ABC with obtuse angle at vertex A as shown in the Pic 1.

Obviously side BC will be the longest side.Β 

Side opposite to largest angle is the longest side in a triangle.

Pic 1
Pic 2

Take a compass and draw an arc of length AB. It will intersect the longest side BC, say at point D.

See Pic 2.

Join AD.

Let us now focus on two trianglesβ–²ABC and β–²ADC

\( \begin{aligned} \overline{AB} &\cong \overline{AD} \ \ (Construction) \\ \overline{AC} &\cong \overline{AC} \ \ (Common) \\ \angle C &= \angle C \ \ (Common) \end{aligned}\)

So we got two congruent sides and one congruent angle in SSA formation but clearly the two triangles are not congruent. Hence, SSA is not a valid congruent criterion.Β 

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]]> Rational Root Theorem – Proof https://mathemafia.com/rational-root-theorem-proof/ Sat, 15 Feb 2025 14:54:11 +0000 https://mathemafia.com/?p=11497 Rational root theorem - Proof. Rational Root Theorem is a very important theorem of Polynomials and is intended for the students learning higher mathematics and for those who are willing to dive deep into the core mathematics. For lower classes, like grade 9 and 10, this theorem comes handy when we have to find the roots of polynomials with degree 3 or more.Β 

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Ex: 8.1, Class 10
Ex: 8.3, Class 10

Rational Root Theorem

Statement: Given a polynomial

𝑃(π‘₯)=π‘Žπ‘›π‘₯𝑛+π‘Žπ‘›βˆ’1π‘₯π‘›βˆ’1+π‘Žπ‘›βˆ’2π‘₯π‘›βˆ’2+…….+π‘Ž1π‘₯+π‘Ž0

with integral coefficients where an β‰  0, a0 β‰  0. if P(x) has a rational root p/q where p and q are relatively prime integers, then p is a divisor of a0 and q is a divisor of an.

Proof

𝑃(π‘₯)=π‘Žπ‘›π‘₯𝑛+π‘Žπ‘›βˆ’1π‘₯π‘›βˆ’1+π‘Žπ‘›βˆ’2π‘₯π‘›βˆ’2+…….+π‘Ž1π‘₯+π‘Ž0π‘€β„Žπ‘’π‘Ÿπ‘’  π‘Ž0,…,π‘Žπ‘›βˆˆβ„€  π‘Žπ‘›π‘‘  π‘Ž0,π‘Žπ‘›β‰ 0.

Let p/q be a rational root of this equation where p and q are coprime integers and q β‰ Β 0.

⇒𝑃(π‘π‘ž)=0β‡’π‘Žπ‘›(π‘π‘ž)𝑛+π‘Žπ‘›βˆ’1(π‘π‘ž)π‘›βˆ’1+π‘Žπ‘›βˆ’2(π‘π‘ž)π‘›βˆ’2+…….+π‘Ž1(π‘π‘ž)+π‘Ž0=0

Multiply both sides by qn

π‘žπ‘›Γ—[π‘Žπ‘›(π‘π‘ž)𝑛+π‘Žπ‘›βˆ’1(π‘π‘ž)π‘›βˆ’1+π‘Žπ‘›βˆ’2(π‘π‘ž)π‘›βˆ’2+…….+π‘Ž2(π‘π‘ž)2+π‘Ž1(π‘π‘ž)+π‘Ž0]=π‘žπ‘›Γ—0π‘Žπ‘›(π‘π‘ž)π‘›π‘žπ‘›+π‘Žπ‘›βˆ’1(π‘π‘ž)π‘›βˆ’1π‘žπ‘›+π‘Žπ‘›βˆ’2(π‘π‘ž)π‘›βˆ’2π‘žπ‘›+…….+π‘Ž2(π‘π‘ž)2π‘žπ‘›+π‘Ž1(π‘π‘ž)π‘žπ‘›+π‘Ž0π‘žπ‘›=0π‘Žπ‘›π‘π‘›+π‘Žπ‘›βˆ’1π‘π‘›βˆ’1π‘ž+π‘Žπ‘›βˆ’2π‘π‘›βˆ’2π‘ž2+…….+π‘Ž2𝑝2π‘žπ‘›βˆ’2+π‘Ž1π‘π‘žπ‘›βˆ’1+π‘Ž0π‘žπ‘›=0  β€¦(1)

Imp: We multiplied by qn to get rid of terms inΒ the denominator.Β 

Video tutorial: Proof of Rational Root TheoremΒ 

Proof of Part 1: p divides a0

Let’s write equation (1) again

π‘Žπ‘›π‘π‘›+π‘Žπ‘›βˆ’1π‘π‘›βˆ’1π‘ž+π‘Žπ‘›βˆ’2π‘π‘›βˆ’2π‘ž2+…….+π‘Ž2𝑝2π‘žπ‘›βˆ’2+π‘Ž1π‘π‘žπ‘›βˆ’1+π‘Ž0π‘žπ‘›=0  β€¦(1)

Take a0qn to the other side of the equation.

π‘Žπ‘›π‘π‘›+π‘Žπ‘›βˆ’1π‘π‘›βˆ’1π‘ž+π‘Žπ‘›βˆ’2π‘π‘›βˆ’2π‘ž2+…….+π‘Ž2𝑝2π‘žπ‘›βˆ’2+π‘Ž1π‘π‘žπ‘›βˆ’1=  β€“π‘Ž0π‘žπ‘›

Now p is common in every term on LHS. This implies:Β 

𝑝[π‘Žπ‘›π‘π‘›βˆ’1+π‘Žπ‘›βˆ’1π‘π‘›βˆ’2π‘ž+π‘Žπ‘›βˆ’2π‘π‘›βˆ’3π‘ž2+…….+π‘Ž2π‘π‘žπ‘›βˆ’2+π‘Ž1π‘žπ‘›βˆ’1]=  β€“π‘Ž0π‘žπ‘›

Now p, q and a0, a1,….an, all are integers. Hence LHS is an integer.

⇒𝑝|𝐿𝐻𝑆⇒𝑝|𝑅𝐻𝑆⇒𝑝|π‘Ž0π‘žπ‘›

As we know that p and q are coprimeΒ 

⇒𝑝  does not divide  π‘žβ‡’𝑝  does not divide  π‘žπ‘›
But because  π‘|π‘Ž0π‘žπ‘›β‡’π‘|π‘Ž0

Proof of Part 2: q divides an

Let’s go back to equation (1)

π‘Žπ‘›π‘π‘›+π‘Žπ‘›βˆ’1π‘π‘›βˆ’1π‘ž+π‘Žπ‘›βˆ’2π‘π‘›βˆ’2π‘ž2+…….+π‘Ž2𝑝2π‘žπ‘›βˆ’2+π‘Ž1π‘π‘žπ‘›βˆ’1+π‘Ž0π‘žπ‘›=0  β€¦(1)

Take anpn to the other side of the equation.

π‘Žπ‘›βˆ’1π‘π‘›βˆ’1π‘ž+π‘Žπ‘›βˆ’2π‘π‘›βˆ’2π‘ž2+…….+π‘Ž2𝑝2π‘žπ‘›βˆ’2+π‘Ž1π‘π‘žπ‘›βˆ’1+π‘Ž0π‘žπ‘›=βˆ’π‘Žπ‘›π‘π‘›  

Now q is common in every term on LHS. This implies:Β 

π‘ž[π‘Žπ‘›βˆ’1π‘π‘›βˆ’1+π‘Žπ‘›βˆ’2π‘π‘›βˆ’2π‘ž+…….+π‘Ž2𝑝2π‘žπ‘›βˆ’3+π‘Ž1π‘π‘žπ‘›βˆ’2+π‘Ž0π‘žπ‘›βˆ’1]=βˆ’π‘Žπ‘›π‘π‘›

Now p, q and a0, a1,….an, all are integers. Hence LHS is an integer.

β‡’π‘ž|πΏπ»π‘†β‡’π‘ž|π‘…π»π‘†β‡’π‘ž|π‘Žπ‘›π‘π‘›

As we know that p and q are coprimeΒ 

β‡’π‘ž  does not divide  π‘β‡’π‘ž  does not divide  π‘π‘›
But because  π‘ž|π‘Žπ‘›π‘π‘›β‡’π‘ž|π‘Žπ‘›

Hope you find this proof useful.

Let’s now use this theorem to find the roots of a polynomial of degree 3.Β 

Example ofΒ Rational Root Theorem

𝑃(π‘₯)=2π‘₯3–3π‘₯2–2π‘₯+3

Find the roots of P(x) using Rational Root Theorem.

This equation is of the form

𝑃(π‘₯)=π‘Ž3π‘₯3+π‘Ž2π‘₯2+π‘Ž1π‘₯+π‘Ž0
β‡’π‘Ž3=2  π‘Žπ‘›π‘‘  π‘Ž0=3

As per rational root theorem, if this polynomial has any rational root, say p/q, then

β‡’  π‘  π‘‘𝑖𝑣𝑖𝑑𝑒𝑠  π‘Ž0  ,  π‘ž  π‘‘𝑖𝑣𝑖𝑑𝑒𝑠  π‘Ž3𝑝  π‘‘𝑖𝑣𝑖𝑑𝑒𝑠  3  ,  π‘ž  π‘‘𝑖𝑣𝑖𝑑𝑒𝑠  2

Here are the possible values of p/q:

π‘π‘žβ‡’π‘π‘ž=Β±1,  Β±3=Β±1,  Β±2=Β±11,Β±31,Β±12,Β±32=Β±1,Β±3,Β±12,Β±32

Now, if P(x) has any rational root, it has to be from the possible values of p/q written above. If P(x) = 0 for any of these values, that will be a rational root of the polynomial. The three values for which P(x) = 0 are 1, -1 and 3/2.

𝑃(π‘₯)𝑃(1)𝑃(βˆ’1)𝑃(32)=2π‘₯3–3π‘₯2–2π‘₯+3=2βˆ’3βˆ’2+3=0=βˆ’2βˆ’3+2+3=0=2Γ—278βˆ’3Γ—94βˆ’2Γ—32+3=274–274βˆ’3+3=0

Rational Root Theorem is a very important theorem of Polynomials and is intended for the students learning higher mathematics and for those who are willing to dive deep into the core mathematics. For lower classes, like class 9 and class 10, this theorem comes handy when we have to find the roots of polynomials with degree 3 or more.Β 

Ex: 8.1, Class 10
Ex: 8.3, Class 10
\(\begin{aligned}Β  \end{aligned}\)

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