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Post a LessonAnswered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Polynomials
Nazia Khanum
What is the number of zeros of the quadratic equation x2+4x+2x2+4x+2?
Answer:
Quadratic Equation: x2+4x+2x2+4x+2
To determine the number of zeros of the quadratic equation, we can use the discriminant method:
Discriminant Formula:
Calculating Discriminant:
Interpreting the Discriminant:
Result:
Conclusion: The number of zeros of the quadratic equation x2+4x+2x2+4x+2 is two.
read lessAnswered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Polynomials
Nazia Khanum
Determining the Value of k
Introduction: To find the value of k when (x – 1) is a factor of the polynomial 4x^3 + 3x^2 – 4x + k, we'll utilize the Factor Theorem.
Factor Theorem: If (x – c) is a factor of a polynomial, then substituting c into the polynomial should result in zero.
Procedure:
Step-by-Step Solution:
Substitute x=1x=1:
Solve for k:
Conclusion: The value of k when (x – 1) is a factor of the given polynomial is k=−3k=−3.
Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Polynomials
Nazia Khanum
Solution: Finding Values of a and b
Given Problem: If x3+ax2–bx+10x3+ax2–bx+10 is divisible by x2–3x+2x2–3x+2, we need to find the values of aa and bb.
Solution Steps:
Step 1: Determine the factors of the divisor
Given divisor: x2–3x+2x2–3x+2
We need to find two numbers that multiply to 22 and add up to −3−3.
The factors of 22 are 11 and 22.
So, the factors that add up to −3−3 are −2−2 and −1−1.
Hence, the divisor factors are (x–2)(x–2) and (x–1)(x–1).
So, the divisor can be written as (x–2)(x–1)(x–2)(x–1).
Step 2: Use Remainder Theorem
If f(x)=x3+ax2–bx+10f(x)=x3+ax2–bx+10 is divisible by (x–2)(x–1)(x–2)(x–1), then the remainder when f(x)f(x) is divided by x2–3x+2x2–3x+2 is zero.
According to Remainder Theorem, if f(x)f(x) is divided by x2–3x+2x2–3x+2, then the remainder is given by f(2)f(2) and f(1)f(1) respectively.
Step 3: Find the value of aa
Substitute x=2x=2 into f(x)f(x) and equate it to 00 to find the value of aa.
f(2)=23+a(2)2–b(2)+10f(2)=23+a(2)2–b(2)+10
0=8+4a–2b+100=8+4a–2b+10
18=4a–2b18=4a–2b
4a–2b=184a–2b=18
Step 4: Find the value of bb
Substitute x=1x=1 into f(x)f(x) and equate it to 00 to find the value of bb.
f(1)=13+a(1)2–b(1)+10f(1)=13+a(1)2–b(1)+10
0=1+a–b+100=1+a–b+10
11=a–b11=a–b
a–b=11a–b=11
Step 5: Solve the equations
Now we have two equations:
We can solve these equations simultaneously to find the values of aa and bb.
Step 6: Solve the equations
Equation 1: 4a–2b=184a–2b=18
Divide by 2: 2a–b=92a–b=9
Equation 2: a–b=11a–b=11
Step 7: Solve the system of equations
Adding equation 2 to equation 1: (2a–b)+(a–b)=9+11(2a–b)+(a–b)=9+11
3a=203a=20
a=203a=320
Substitute a=203a=320 into equation 2: 203–b=11320–b=11
b=203–11b=320–11
b=20–333b=320–33
b=−133b=3−13
Step 8: Final values of aa and bb
a=203a=320
b=−133b=3−13
So, the values of aa and bb are a=203a=320 and b=−133b=3−13 respectively.
Answered on 18 Apr Learn CBSE/Class 9/Mathematics/Algebra/Polynomials
Nazia Khanum
Monomial and Binomial Examples with Degrees
Monomial Example (Degree: 82)
Binomial Example (Degree: 99)
Additional Notes:
Answered on 02 Feb Learn CBSE/Class 9/Mathematics/Algebra/Polynomials
Pooja R. Jain
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