Answer :
To solve the quadratic equation [tex]\(4x^2 + 8x + 3 = 0\)[/tex], we can use the quadratic formula. The formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients from the equation are:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = 8 \)[/tex]
- [tex]\( c = 3 \)[/tex]
First, we calculate the discriminant, [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ b^2 = 8^2 = 64 \][/tex]
[tex]\[ 4ac = 4 \times 4 \times 3 = 48 \][/tex]
So, the discriminant is:
[tex]\[ 64 - 48 = 16 \][/tex]
Since the discriminant is positive, the equation has two real and distinct solutions.
Now, we can find the values of [tex]\( x \)[/tex] using the quadratic formula:
1. Find [tex]\(\frac{-b + \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x_1 = \frac{-8 + \sqrt{16}}{8} = \frac{-8 + 4}{8} = \frac{-4}{8} = -0.5 \][/tex]
2. Find [tex]\(\frac{-b - \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x_2 = \frac{-8 - \sqrt{16}}{8} = \frac{-8 - 4}{8} = \frac{-12}{8} = -1.5 \][/tex]
Therefore, the solutions are [tex]\( x = -0.5 \)[/tex] and [tex]\( x = -1.5 \)[/tex].
The correct option is:
(D) [tex]\( x = -1.5 \)[/tex]; [tex]\( x = -0.5 \)[/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients from the equation are:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = 8 \)[/tex]
- [tex]\( c = 3 \)[/tex]
First, we calculate the discriminant, [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ b^2 = 8^2 = 64 \][/tex]
[tex]\[ 4ac = 4 \times 4 \times 3 = 48 \][/tex]
So, the discriminant is:
[tex]\[ 64 - 48 = 16 \][/tex]
Since the discriminant is positive, the equation has two real and distinct solutions.
Now, we can find the values of [tex]\( x \)[/tex] using the quadratic formula:
1. Find [tex]\(\frac{-b + \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x_1 = \frac{-8 + \sqrt{16}}{8} = \frac{-8 + 4}{8} = \frac{-4}{8} = -0.5 \][/tex]
2. Find [tex]\(\frac{-b - \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x_2 = \frac{-8 - \sqrt{16}}{8} = \frac{-8 - 4}{8} = \frac{-12}{8} = -1.5 \][/tex]
Therefore, the solutions are [tex]\( x = -0.5 \)[/tex] and [tex]\( x = -1.5 \)[/tex].
The correct option is:
(D) [tex]\( x = -1.5 \)[/tex]; [tex]\( x = -0.5 \)[/tex]