Math Problem Statement

Three math problems related to quadratic equations, as shown in the uploaded image.

Solution

Let's solve the questions in the uploaded image one by one.

Question 46:

If α,β\alpha, \beta are the roots of x2+ax−b=0x^2 + ax - b = 0 and γ,δ\gamma, \delta are the roots of x2+ax+b=0x^2 + ax + b = 0, then (α−γ)(α−δ)(β−γ)(β−δ)(\alpha - \gamma)(\alpha - \delta)(\beta - \gamma)(\beta - \delta) is:

Solution:

The roots are as follows:

  1. For x2+ax−b=0x^2 + ax - b = 0: The sum of roots α+β=−a\alpha + \beta = -a and the product αβ=−b\alpha\beta = -b.
  2. For x2+ax+b=0x^2 + ax + b = 0: The sum of roots γ+δ=−a\gamma + \delta = -a and the product γδ=b\gamma\delta = b.

Now, we simplify (α−γ)(α−δ)(β−γ)(β−δ)(\alpha - \gamma)(\alpha - \delta)(\beta - \gamma)(\beta - \delta):

= (\alpha\beta - \alpha(\gamma + \delta) + \gamma\delta)(\beta - \gamma)(\beta - \delta).$$ After simplifying (and substituting the relations between the roots), the expression evaluates to: $$4b^2.$$ Thus, the answer is **(d) 4b²**. --- ### **Question 47:** If the roots of $$4x^2 + 5k = (5k + 1)x$$ differ by unity, then the negative value of $$k$$ is: #### **Solution:** The equation is rewritten as: $$4x^2 - (5k + 1)x + 5k = 0.$$ Let the roots be $$p$$ and $$p+1$$. By Vieta's formulas: 1. Sum of roots: $$p + (p+1) = \frac{-(5k + 1)}{4}.$$ Simplify: $$2p + 1 = \frac{-(5k + 1)}{4}.$$ 2. Product of roots: $$p(p+1) = \frac{5k}{4}.$$ Expand: $$p^2 + p = \frac{5k}{4}.$$ Solve these equations simultaneously to find $$k$$. After solving, the negative value of $$k$$ is: $$\boxed{-\frac{3}{5}}.$$ Thus, the answer is **(d) $$-\frac{3}{5}$$**. --- ### **Question 48:** If $$\alpha, \beta$$ are the roots of $$2x^2 - 35x + 2 = 0$$, the value of $$(2\alpha - 35)^3(2\beta - 35)^3$$ is: #### **Solution:** Given $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation: $$2x^2 - 35x + 2 = 0.$$ From Vieta's formulas: 1. Sum of roots: $$\alpha + \beta = \frac{-(-35)}{2} = \frac{35}{2}.$$ 2. Product of roots: $$\alpha\beta = \frac{2}{2} = 1.$$ Now, evaluate $$(2\alpha - 35)(2\beta - 35)$$: $$(2\alpha - 35)(2\beta - 35) = 4\alpha\beta - 70(\alpha + \beta) + 1225.$$ Substitute the values: $$(2\alpha - 35)(2\beta - 35) = 4(1) - 70\left(\frac{35}{2}\right) + 1225.$$ Simplify: $$(2\alpha - 35)(2\beta - 35) = 4 - 1225 + 1225 = 4.$$ Thus: $$(2\alpha - 35)^3(2\beta - 35)^3 = 4^3 = 64.$$ The answer is **(b) 64**. --- Let me know if you need detailed steps for any part! --- ### **5 Related Questions:** 1. How can we verify the roots of a quadratic equation are correct? 2. What are the implications of Vieta's formulas in solving polynomial equations? 3. How does the product of roots help in deriving key expressions in quadratic problems? 4. How can one simplify higher powers in polynomial root-related problems like Question 48? 5. What happens if roots differ by some arbitrary constant instead of unity? --- ### **Tip:** Always use Vieta's formulas to simplify root-based expressions and solve equations efficiently.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Quadratic equations
Roots of equations
Vieta's formulas
Expression simplification
Polynomial relationships

Formulas

Sum of roots: α + β = -b/a
Product of roots: αβ = c/a
Expression evaluation with roots
Power and product rules in polynomials

Theorems

Vieta's Theorem
Properties of quadratic equations

Suitable Grade Level

Grades 11-12

Related Recommendation

Solving Polynomial Equations: Vieta's Formulas and Factorization

Solving Polynomial and Quadratic Equations with Root Proportions and Inequalities

Solve Quadratic Equations and Simplify Square Roots - Step-by-Step Solutions

Quadratic Equations: Solving Roots and Transforming Equations

Solving Quadratic Equations with Complex Roots