- This topic has 0 replies, 1 voice, and was last updated 1 month, 1 week ago by .
Viewing 1 post (of 1 total)
Viewing 1 post (of 1 total)
- You must be logged in to reply to this topic.
No products in the cart.
Tagged: quadratic equation
Given that one root of equation $x^2+px +12=0$ is 4 find the value of p and other root.
### Step 1: Finding the value of \(p\) using the root \(x = 4\)
Given \(x^2 + px + 12 = 0\) and one root \(x = 4\), substituting \(x = 4\) gives us:
\[
4^2 + 4p + 12 = 0
\]
\[
16 + 4p + 12 = 0
\]
\[
28 + 4p = 0
\]
\[
4p = -28
\]
\[
p = -7
\]
### Step 2: Finding the other root
With \(p = -7\), our equation becomes \(x^2 – 7x + 12 = 0\).
Given that the sum of the roots (\(x_1 + x_2\)) of a quadratic equation \(ax^2 + bx + c = 0\) is \(-\frac{b}{a}\), and the product (\(x_1 \cdot x_2\)) is \(\frac{c}{a}\), we can find the other root without solving the equation again.
Since one root is \(4\), and we know \(p = -7\), we can find the other root by:
Sum of the roots = \(x_1 + x_2 = -\frac{b}{a} = -\frac{-7}{1} = 7\)
Thus, \(4 + x_2 = 7\)
So, \(x_2 = 3\).
Therefore, with one root being \(4\), and the value of \(p\) is \(-7\), the other root of the equation is \(3\).
© 2023 Created with Royal Elementor Addons