Question

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

(2) -1, 4

Step-by-step solution

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: \[ x^2 + b x + q = 0. \].

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 \[ q = 2. \]

Form the Equation with Given Roots

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

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: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} \] \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \] \[ x = \frac{3 \pm \sqrt{25}}{2} \] \[ x = \frac{3 \pm 5}{2} \].

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.

Key concepts

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’s positive, we get two distinct real roots. If it’s zero, we get one real root (a repeated root). If it’s 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.