Viewing 1 post (of 1 total)
• Author
Posts
• #24374
Keymaster

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 $$3$$.

Viewing 1 post (of 1 total)
• You must be logged in to reply to this topic.