Find the quadratic polynomial the sum and product of its zeros are -1 by (means ÷ ) 4 & 1 by ( means ÷ ) 3
Sum of class 10th
Don't give wrong answer please
and one request is there please solve this question step by step with explanation
Answers
Step-by-step explanation:
Answer:-
According to the Question
It is given that ,
Sum of zeros = -1/4
Product of zeros = 1/3
We have to calculate the quadratic polynomial .
As we know that quadratic polynomial when the sum and product of its zeros are given by
f(x) = x² -(sum of zeros)x + product of zeros
Substitute the value we get
→ f(x) = x² -(-1/4)x + 1/3
→ f(x) = x² + (1/4)x + 1/3
So, the required quadratic polynomial is
f(x) = x² + (1/4)x + 1/3
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Extra Information !!
• (a + b)² = a² + b² +2ab
• ( a - b )² = a² + b² -2ab
• ( a² - b² ) = ( a - b ) ( a + b )
• ( a +b +c)² = (a² + b² + c²) + 2(ab + bc +ca)
• (a + b)³ = a³ + b³ + 3ab(a+b)
• (a-b)³ = a³ - b³ -3ab(a-b)
• ( a³ + b³) = (a+b) (a²-ab +b²)
• (a³ - b³) = (a -b) (a² + ab + b²)
Answer:
Answer:
Given :-
The resístance of each resístors is 10 ohm resistors.
To Find :-
What is the maximum resístance make available using two resístors.
Formula Used :-
\clubsuit♣ Equívalent Resístance for series connection :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{R_{eq} =\: R_1 + R_2\: +\: . . . .\: +\: R_n}}}\\\end{gathered}↦Req=R1+R2+....+Rn
where,
\sf R_{eq}Req = Equívalent Resístance
\sf R_1R1 = Resístance of resístors R₁
\sf R_2R2 = Resístance of resístors R₂
Solution :-
Given :
\begin{gathered}\bigstar\: \rm{\bold{Resistance\: of\: resistors\: (R_1) =\: 10\: \text{\O}mega}}\\\end{gathered}★Resistanceofresistors(R1)=10Ømega
\bigstar\: \rm{\bold{Resistance\: of\: resistors\: (R_2) =\: 10\: \text{\O}mega}}★Resistanceofresistors(R2)=10Ømega
According to the question by using the formula we get,
\longrightarrow \sf R_{eq} =\: R_1 + R_2⟶Req=R1+R2
\longrightarrow \sf R_{eq} =\: 10\: \text{\O}mega + 10\: \text{\O}mega⟶Req=10Ømega+10Ømega
\longrightarrow \sf R_{eq} =\: (10 + 10)\: \text{\O}mega⟶Req=(10+10)Ømega
\longrightarrow \sf R_{eq} =\: (20)\: \text{\O}mega⟶Req=(20)Ømega
\longrightarrow \sf\bold{\red{R_{eq} =\: 20\: \text{\O}mega}}⟶Req=20Ømega
\therefore∴ The maximum resístance is 20 Ω . It is connected by equívalent resístance of series connection.
\begin{gathered}\\\end{gathered}
EXTRA INFORMATION :-
\clubsuit♣ Equívalent Resístance for párallel connection :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{R_{eq}} =\: \dfrac{1}{R_1} + \dfrac{1}{R_2}\: +\: . . . .\: +\: \dfrac{1}{R_n}}}}\\\end{gathered}↦Req1=R11+R21+....+Rn1
where,
\sf R_{eq}Req = Equívalent Resístance
\sf R_1R1 = Resístance of resístors R₁
\sf R_2R2 = Resístance of resístors R₂