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If the sum and product of the zeros of a quadratic polynomial are 3 and -10, respectively, then the quadratic polynomial is:
Step 1: Understanding Zeros and Quadratic Form - A quadratic polynomial has the general form: x² + bx + c - Its zeros are the roots that make the polynomial equal to zero - Let the zeros be p and q Step 2: Given Conditions - Sum of zeros (p + q) = 3 - Product of zeros (p * q) = -10 Step 3: RelationsRead more
Step 1: Understanding Zeros and Quadratic Form
– A quadratic polynomial has the general form: x² + bx + c
– Its zeros are the roots that make the polynomial equal to zero
– Let the zeros be p and q
Step 2: Given Conditions
– Sum of zeros (p + q) = 3
– Product of zeros (p * q) = -10
Step 3: Relationship to Quadratic Coefficients
In the standard form x² + bx + c:
– b = -(sum of zeros)
– c = product of zeros
Step 4: Calculating Coefficients
– b = -(p + q) = -3
– c = p * q = -10
Step 5: Constructing the Quadratic Polynomial
The polynomial becomes:
x² – 3x – 10
Verification:
– Coefficient of x²: 1
– Coefficient of x: -3 (negative of zero sum)
– Constant term: -10 (product of zeros)
Mathematical Reasoning:
The coefficients directly reflect the given conditions about the zeros:
– Sum of zeros: p + q = 3
– Product of zeros: p * q = -10
Conclusion:
The quadratic polynomial is x² – 3x – 10.
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The polynomial x³ + 3x² + 3x + 1 is divisible by:
Step 1: Polynomial Structure Given polynomial: f(x) = x³ + 3x² + 3x + 1 Step 2: Divisibility Test Method - To check if a polynomial is divisible by (x + a), we can use the remainder theorem - The polynomial is divisible by (x + a) if f(-a) = 0 Step 3: Checking Possible Divisors Let's evaluate f(-1):Read more
Step 1: Polynomial Structure
Given polynomial: f(x) = x³ + 3x² + 3x + 1
Step 2: Divisibility Test Method
– To check if a polynomial is divisible by (x + a), we can use the remainder theorem
– The polynomial is divisible by (x + a) if f(-a) = 0
Step 3: Checking Possible Divisors
Let’s evaluate f(-1):
f(-1) = (-1)³ + 3(-1)² + 3(-1) + 1
= -1 + 3 – 3 + 1
= 0
Mathematical Insight:
– When f(-1) = 0, it means (x + 1) is a factor of the polynomial
– This indicates the polynomial is completely divisible by (x + 1)
Step 4: Verification
Dividing x³ + 3x² + 3x + 1 by (x + 1):
– The result is a quadratic: x² + 2x + 1
– (x + 1)(x² + 2x + 1) = x³ + 3x² + 3x + 1 ✓
Conclusion:
The polynomial is divisible by x + 1.
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If α and β are the zeros of the polynomial x² – 5x + 6, then the value of α² + β² is:
Step 1: Understanding the Polynomial Polynomial: x² - 5x + 6 - Zeros are α and β - Standard form: x² - (α + β)x + (α * β) = 0 Step 2: Vieta's Formulas We already know from the given polynomial: - Sum of zeros: α + β = 5 - Product of zeros: α * β = 6 Step 3: Goal Calculation We are trying to calculatRead more
Step 1: Understanding the Polynomial
Polynomial: x² – 5x + 6
– Zeros are α and β
– Standard form: x² – (α + β)x + (α * β) = 0
Step 2: Vieta’s Formulas
We already know from the given polynomial:
– Sum of zeros: α + β = 5
– Product of zeros: α * β = 6
Step 3: Goal Calculation
We are trying to calculate: α² + β²
Step 4: Algebraic Manipulation
(α + β)² = α² + 2αβ + β²
α² + β² = (α + β)² – 2αβ
Step 5: Substitution
– (α + β)² = 5²
= 25
– αβ = 6
– α² + β² = (α + β)² – 2(αβ)
= 25 – 12
= 11
Mathematical Insight
This approach applies Vieta’s formulas to connect the coefficients of a quadratic with the characteristics of its zeros without actually solving for the zeros themselves.
Conclusion:
The value of α² + β² is 11.
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If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =
Step 1: Understanding Infinitely Many Solutions • In a system of linear equations, infinitely many solutions occur when the equations represent the same line. • This means the equations are scalar multiples of each other. Step 2: Mathematical Representation Given equations: 1. 2x + 3y = 5 (EquationRead more
Step 1: Understanding Infinitely Many Solutions
• In a system of linear equations, infinitely many solutions occur when the equations represent the same line.
• This means the equations are scalar multiples of each other.
Step 2: Mathematical Representation
Given equations:
1. 2x + 3y = 5 (Equation ₁)
2. 4x + ky = 10 (Equation ₂)
Step 3: Condition for Infinitely Many Solutions
For the lines to be the same, the coefficients must maintain a consistent ratio.
Coefficient Analysis:
• 1st equation: 2x + 3y = 5
– Coefficient of x: 2
– Coefficient of y: 3
• 2nd equation: 4x + ky = 10
– Coefficient of x: 4
– Coefficient of y: k
Step 4: Ratio Consistency
Coefficient of x ratio: 4 ÷ 2 = 2
Coefficient of y ratio: k ÷ 3 = 2
Therefore:
k = 3 * 2 = 6
Step 5: Verification
• When k = 6, the second equation becomes: 4x + 6y = 10
• Dividing by 2: 2x + 3y = 5 (exactly the same as the first equation)
Conclusion:
The value of k must be 6 to create infinitely many solutions.
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The value of k for which the pair of equations 3x – y = 5 and 6x – 2y = k has infinitely many solutions is:
Step 1: Examining the System of Equations Equation ₁: 3x - y = 5 Equation ₂: 6x - 2y = k Step 2: Infinitely Many Solutions Condition Infinitely many solutions are achieved when the equations define the same line. That is, the equations should be scalar multiples of one another. Step 3: Coefficient CRead more
Step 1: Examining the System of Equations
Equation ₁: 3x – y = 5
Equation ₂: 6x – 2y = k
Step 2: Infinitely Many Solutions Condition
Infinitely many solutions are achieved when the equations define the same line.
That is, the equations should be scalar multiples of one another.
Step 3: Coefficient Comparison
– Equation ₁ coefficients:
– x coefficient: 3
– y coefficient: -1
– Equation ₂ coefficients:
– x coefficient: 6
– y coefficient: -2
Step 4: Checking Proportionality
Notice the proportionality of coefficients:
– x coefficient ratio: 6 ÷ 3 = 2
– y coefficient ratio: -2 ÷ (-1) = 2
Step 5: Constant Term Condition
For an infinite number of solutions, constant terms must also be under the same scaling.
Equations should be equal:
3x – y = 5
6x – 2y = k
Replacing the coefficient scaling:
– If the equations are for the same line, k should be 2 * 5 = 10
Step 6: Checking
If k = 10, the second equation is:
6x – 2y = 10
Divide by 2:
3x – y = 5 (Which is exactly the same as the first equation)
Conclusion:
The value of k for infinitely many solutions is 10.
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