Question 7

The zeroes of the quadratic polynomial $x^2+99x+127$ are

Solution:

To find the zeroes of the quadratic polynomial $$x^2+99x+127$$, we will follow these steps:

Step 1: Write the polynomial and set it equal to zero
We start with the polynomial:
$$x^2+99x+127=0$$

Step 2: Identify coefficients
From the polynomial, we identify the coefficients:
– $$a=1$$ (coefficient of $${\mathrm{x2}}^{}$$)
– $$b=99$$ (coefficient of $$x$$)
– $$c=127$$ (constant term)

Step 3: Calculate the discriminant
The discriminant $$D$$ is given by the formula:
$$D={\mathrm{b2}}^{}-4ac$$
Substituting the values of $$a$$, $$b$$, and $$c$$:
$$D={992}^{}-4\cdot 1\cdot 127$$
Calculating $${99}^2$$:
$${992}^{}=9801$$
Calculating $$4\cdot 1\cdot 127$$:
$$4\cdot 127=508$$
Now substituting these values back into the discriminant formula:
$$D=9801-508=9293$$

Step 4: Calculate the square root of the discriminant
Next, we find the square root of the discriminant:
$$\sqrt{D}=\sqrt{9293}\approx 96.4$$

Step 5: Use the quadratic formula to find the zeroes
The zeroes of the polynomial can be found using the quadratic formula:
$$x=(\frac{-}{b}$$
Substituting the values of $$b$$, $$\sqrt{D}$$, and $$a$$:
$$x=(\frac{-}{99}$$
This gives us two equations to solve:
1. $$x_1=(\frac{-}{99}$$
2. $$x_2=(\frac{-}{99}$$

Step 6: Calculate the first zero
Calculating $${\mathrm{x1}}^{}$$:
$${\mathrm{x1}}^{}=(-99+96.4)/2=\frac{-}{2.6}=-1.3$$

Step 7: Calculate the second zero
Calculating $${\mathrm{x2}}^{}$$:
$$x2=(-99-96.4)/2=-97.7$$

Conclusion
The zeroes of the quadratic polynomial $$x^2+99x+127$$ are:
$${\mathrm{x1}}^{}=-1.3\text{and}{\mathrm{x2}}^{}=-97.7$$