Answer :
Alright, let's solve the equation step-by-step!
We start with the equation:
[tex]\[ \left(2 a^2 + 5 a + 2\right)^{\frac{1}{2}} = 3 \][/tex]
To eliminate the square root, we can square both sides of the equation:
[tex]\[ \left[ \left(2 a^2 + 5 a + 2\right)^{\frac{1}{2}} \right]^2 = 3^2 \][/tex]
This simplifies to:
[tex]\[ 2 a^2 + 5 a + 2 = 9 \][/tex]
Next, we subtract 9 from both sides to set the equation to 0:
[tex]\[ 2 a^2 + 5 a + 2 - 9 = 0 \][/tex]
So, we have:
[tex]\[ 2 a^2 + 5 a - 7 = 0 \][/tex]
This is a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex].
To find the solutions for [tex]\(a\)[/tex], we can use the quadratic formula:
[tex]\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -7\)[/tex].
Plugging in these values:
[tex]\[ a = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2} \][/tex]
Simplifying inside the square root:
[tex]\[ a = \frac{-5 \pm \sqrt{25 + 56}}{4} \][/tex]
[tex]\[ a = \frac{-5 \pm \sqrt{81}}{4} \][/tex]
Since [tex]\(\sqrt{81} = 9\)[/tex], we have:
[tex]\[ a = \frac{-5 \pm 9}{4} \][/tex]
Now, we solve for the two possible values of [tex]\(a\)[/tex]:
1. For [tex]\(a = \frac{-5 + 9}{4}\)[/tex]:
[tex]\[ a = \frac{4}{4} = 1 \][/tex]
2. For [tex]\(a = \frac{-5 - 9}{4}\)[/tex]:
[tex]\[ a = \frac{-14}{4} = -3.5 \][/tex]
Therefore, the solutions to the equation are:
[tex]\[ a = 1 \quad \text{or} \quad a = -3.5 \][/tex]
Hence, the correct answers are:
[tex]\[ a = -\frac{7}{2} \quad \text{(which is -3.5)} \quad \text{or} \quad a = 1 \][/tex]
We start with the equation:
[tex]\[ \left(2 a^2 + 5 a + 2\right)^{\frac{1}{2}} = 3 \][/tex]
To eliminate the square root, we can square both sides of the equation:
[tex]\[ \left[ \left(2 a^2 + 5 a + 2\right)^{\frac{1}{2}} \right]^2 = 3^2 \][/tex]
This simplifies to:
[tex]\[ 2 a^2 + 5 a + 2 = 9 \][/tex]
Next, we subtract 9 from both sides to set the equation to 0:
[tex]\[ 2 a^2 + 5 a + 2 - 9 = 0 \][/tex]
So, we have:
[tex]\[ 2 a^2 + 5 a - 7 = 0 \][/tex]
This is a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex].
To find the solutions for [tex]\(a\)[/tex], we can use the quadratic formula:
[tex]\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -7\)[/tex].
Plugging in these values:
[tex]\[ a = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2} \][/tex]
Simplifying inside the square root:
[tex]\[ a = \frac{-5 \pm \sqrt{25 + 56}}{4} \][/tex]
[tex]\[ a = \frac{-5 \pm \sqrt{81}}{4} \][/tex]
Since [tex]\(\sqrt{81} = 9\)[/tex], we have:
[tex]\[ a = \frac{-5 \pm 9}{4} \][/tex]
Now, we solve for the two possible values of [tex]\(a\)[/tex]:
1. For [tex]\(a = \frac{-5 + 9}{4}\)[/tex]:
[tex]\[ a = \frac{4}{4} = 1 \][/tex]
2. For [tex]\(a = \frac{-5 - 9}{4}\)[/tex]:
[tex]\[ a = \frac{-14}{4} = -3.5 \][/tex]
Therefore, the solutions to the equation are:
[tex]\[ a = 1 \quad \text{or} \quad a = -3.5 \][/tex]
Hence, the correct answers are:
[tex]\[ a = -\frac{7}{2} \quad \text{(which is -3.5)} \quad \text{or} \quad a = 1 \][/tex]
