Answer :
To solve the quadratic equation [tex]\( x^2 + 10x + 24 = 0 \)[/tex], we can use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 10 \)[/tex]
- [tex]\( c = 24 \)[/tex]
Step-by-step solution:
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = 10^2 - 4 \cdot 1 \cdot 24 = 100 - 96 = 4 \][/tex]
2. Find the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{4} = 2 \][/tex]
3. Apply the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-10 \pm 2}{2 \cdot 1} \][/tex]
4. Calculate the two possible solutions:
- For the positive root:
[tex]\[ x_1 = \frac{-10 + 2}{2} = \frac{-8}{2} = -4.0 \][/tex]
- For the negative root:
[tex]\[ x_2 = \frac{-10 - 2}{2} = \frac{-12}{2} = -6.0 \][/tex]
Therefore, the solutions to the equation [tex]\( x^2 + 10x + 24 = 0 \)[/tex] are:
[tex]\[ x_1 = -4.0 \][/tex]
[tex]\[ x_2 = -6.0 \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 10 \)[/tex]
- [tex]\( c = 24 \)[/tex]
Step-by-step solution:
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = 10^2 - 4 \cdot 1 \cdot 24 = 100 - 96 = 4 \][/tex]
2. Find the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{4} = 2 \][/tex]
3. Apply the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-10 \pm 2}{2 \cdot 1} \][/tex]
4. Calculate the two possible solutions:
- For the positive root:
[tex]\[ x_1 = \frac{-10 + 2}{2} = \frac{-8}{2} = -4.0 \][/tex]
- For the negative root:
[tex]\[ x_2 = \frac{-10 - 2}{2} = \frac{-12}{2} = -6.0 \][/tex]
Therefore, the solutions to the equation [tex]\( x^2 + 10x + 24 = 0 \)[/tex] are:
[tex]\[ x_1 = -4.0 \][/tex]
[tex]\[ x_2 = -6.0 \][/tex]