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Chapter 7: Problem 4

If the roots of the equation \(x^{2}+p x+c=0\) are \(2,-2\) and the roots of the equation \(x^{2}+b x+q=0\) are \(-1,-2\), then the roots of the equation \(x^{2}+b x+c=0\) are \((1)-3,-2\) \((2)-1,4\) (3) \(1,-4\) (4) \(-2,3\)

Short Answer

Expert verified
(2) -1, 4

Step by step solution

01

Identify the Given Roots and Equations

Identify the given roots for the two quadratic equations: Equation 1: Roots are 2 and -2, equation is: \[ x^2 + p x + c = 0. \] Equation 2: Roots are -1 and -2, equation is: \[ x^2 + b x + q = 0. \].
02

Apply Vieta's Formulas

Use Vieta's formulas to determine the coefficients p, c, b, and q. For the equation with roots 2 and -2:Sum of the roots \((2) + (-2) = 0\) so, \[ p = 0. \] Product of the roots \((2) \times (-2) = -4 \)so, \[ c = -4. \] For the equation with roots -1 and -2:Sum of the roots \((-1) + (-2) = -3\), \[ b = -3. \] Product of the roots \((-1) \times (-2) = 2\), so \[ q = 2. \]
03

Form the Equation with Given Roots

Now, form the equation using the determined values of b and c. The equation is: \[ x^2 - 3x - 4 = 0.\]
04

Determine Possible Roots

To find the roots of the equation \( x^2 - 3x - 4 = 0 \), use the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1, b = -3, \text{and } c = -4 \). Substitute the values: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} \] \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \] \[ x = \frac{3 \pm \sqrt{25}}{2} \] \[ x = \frac{3 \pm 5}{2} \].
05

Solve for Roots

Solve the equations to find the roots: \[ x = \frac{3 + 5}{2} = 4 \] and \[ x = \frac{3 - 5}{2} = -1\]. Therefore, the roots of the equation \( x^2 + bx + c = 0 \) are -1 and 4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vieta's Formulas
Vieta's formulas offer an excellent way to connect the coefficients of a polynomial to the sum and product of its roots. When dealing with a quadratic equation of the form \(\text{ax}^2 + \text{bx} + \text{c} = 0\), Vieta's formulas state that:
  • The sum of the roots (\(\text{r}_1 + \text{r}_2\)) is equal to -\(b/a\).
  • The product of the roots (\(\text{r}_1 \times \text{r}_2\)) is equal to \(c/a\).

In our case, we can use these to simplify our problem significantly. For instance, if the roots are specified, such as 2 and -2, we can easily determine the coefficients of the quadratic equation by substituting these values into Vieta's formulas.
Roots of Quadratic Equations
Roots of a quadratic equation are solutions for \(x\) that satisfy the equation \(\text{ax}^2 + \text{bx} + \text{c} = 0\). These roots can be real or complex numbers. When dealing with roots, we look for values of \(x\) that make the equation true.
For example, in the exercise:
  • Roots for \(x^2 + px + c = 0\) are 2 and -2. This directly tells us that \(p = 0\) and \(c = -4\).
  • For \(x^2 + bx + q = 0\), the roots are -1 and -2, hence \(b = -3\) and \(q = 2\).

The key idea is to understand that once you have the roots, determining the coefficients is straightforward using Vieta's formulas.
Quadratic Formula
The quadratic formula is a universal method for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is:
\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\.
  • The term under the square root, \(b^2 - 4ac\), is known as the discriminant.
  • The discriminant helps us understand the nature of the roots. If it鈥檚 positive, we get two distinct real roots. If it鈥檚 zero, we get one real root (a repeated root). If it鈥檚 negative, we get complex roots.

For the given problem equation \(x^2 - 3x - 4 = 0\), you can find the roots by plugging \(a=1\), \(b=-3\), and \(c=-4\) into the quadratic formula:
\ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} = \frac{3 \pm 5}{2} \.
This gives us the roots 4 and -1, which matches our requirement perfectly. This method is incredibly reliable and a must-master tool for tackling quadratic equations.

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Most popular questions from this chapter

A ring of radius \(R\) is having two charges \(q\) and \(2 q\) distributed on its two half parts. The electric potential at a point on its axis at a distance \(2 \sqrt{2} \mathrm{R}\) from its centre is : (1) \(\frac{3 \mathrm{kq}}{\mathrm{R}}\) (2) \(\frac{\mathrm{kq}}{3 \mathrm{R}}\) (3) \(\frac{\mathrm{kq}}{\mathrm{R}}\) (4) \(\frac{\mathrm{kq}}{\sqrt{3 \mathrm{R}}}\)

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