Find the values of x if p + 7 = 0, q - 12 = 0 and x^2 + px + q = 0. If x = p is a solution of the equation x(2x + 5) = 3, then find the values of p. If x = 3 is a solution of the e... Find the values of x if p + 7 = 0, q - 12 = 0 and x^2 + px + q = 0. If x = p is a solution of the equation x(2x + 5) = 3, then find the values of p. If x = 3 is a solution of the equation (k + 2)x^2 - kx + 6 = 0, find the value of k. Hence, find the other root of the equation. Read more

Steps to Solve

1. Find the value of ( p ) using the first equation

We start from the equation ( x(2x + 5) = 3 ). First, we set this equation equal to zero:
$$ x(2x + 5) - 3 = 0 $$
This leads to:
$$ 2x^2 + 5x - 3 = 0 $$

2. Apply the quadratic formula

To solve for ( x ), we can use the quadratic formula, which is given by:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
For ( 2x^2 + 5x - 3 = 0 ), we have:

• ( a = 2 )
• ( b = 5 )
• ( c = -3 )

Plugging these into the quadratic formula, we calculate:
$$ b^2 - 4ac = 5^2 - 4(2)(-3) = 25 + 24 = 49 $$
Now substituting back:
$$ x = \frac{-5 \pm \sqrt{49}}{2(2)} = \frac{-5 \pm 7}{4} $$
This gives two values:
$$ x_1 = \frac{2}{4} = \frac{1}{2}, \quad x_2 = \frac{-12}{4} = -3 $$

Since ( x = p ), we conclude that: $$ p = \frac{1}{2} \text{ or } p = -3 $$

3. Finding the value of ( k )

Now, for the second part, we need to substitute ( x = 3 ) into the equation ( (k + 2)x^2 - kx + 6 = 0 ):
$$ (k + 2)(3^2) - k(3) + 6 = 0 $$
This simplifies to:
$$ 9(k + 2) - 3k + 6 = 0 $$
Expanding:
$$ 9k + 18 - 3k + 6 = 0 $$
Combining like terms:
$$ 6k + 24 = 0 $$
Solving for ( k ):
$$ 6k = -24 $$
Thus,
$$ k = -4 $$

4. Find the other root of the quadratic equation

Using the first root ( x = 3 ) and ( k = -4 ), we can now find the second root by applying Vieta's formulas, which tells us that for a quadratic of the form ( ax^2 + bx + c = 0 ), the sum of the roots ( r_1 + r_2 ) is given by ( -\frac{b}{a} ).
In our case:

• ( a = k + 2 = -4 + 2 = -2 )
• ( b = -k = 4 )
  So:
  $$ r_1 + r_2 = -\frac{4}{-2} = 2 $$
  Knowing one root ( r_1 = 3 ):
  $$ 3 + r_2 = 2 $$ Thus,
  $$ r_2 = 2 - 3 = -1 $$