Answer :
To solve the quadratic equation [tex]\(2x^2 + 2x = 9\)[/tex], we first need to rewrite it in standard form [tex]\(ax^2 + bx + c = 0\)[/tex].
Starting from:
[tex]\[ 2x^2 + 2x = 9 \][/tex]
Subtract 9 from both sides to obtain:
[tex]\[ 2x^2 + 2x - 9 = 0 \][/tex]
Here, the coefficients are [tex]\(a = 2\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -9\)[/tex].
We will use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, calculate the discriminant, [tex]\(\Delta\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 2^2 - 4(2)(-9) \][/tex]
[tex]\[ \Delta = 4 + 72 \][/tex]
[tex]\[ \Delta = 76 \][/tex]
Since the discriminant is positive, we will have two distinct real roots.
Next, we find the two solutions using the quadratic formula:
For [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x_1 = \frac{-2 + \sqrt{76}}{2(2)} \][/tex]
[tex]\[ x_1 = \frac{-2 + \sqrt{76}}{4} \][/tex]
[tex]\[ x_1 \approx \frac{-2 + 8.72}{4} \][/tex]
[tex]\[ x_1 \approx \frac{6.72}{4} \][/tex]
[tex]\[ x_1 \approx 1.68 \][/tex]
For [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x_2 = \frac{-2 - \sqrt{76}}{2(2)} \][/tex]
[tex]\[ x_2 = \frac{-2 - 8.72}{4} \][/tex]
[tex]\[ x_2 \approx \frac{-10.72}{4} \][/tex]
[tex]\[ x_2 \approx -2.68 \][/tex]
Thus, the solutions to the quadratic equation [tex]\(2x^2 + 2x - 9 = 0\)[/tex] are:
[tex]\[ x \approx 1.68, -2.68 \][/tex]
These results are rounded to the nearest hundredth.
Starting from:
[tex]\[ 2x^2 + 2x = 9 \][/tex]
Subtract 9 from both sides to obtain:
[tex]\[ 2x^2 + 2x - 9 = 0 \][/tex]
Here, the coefficients are [tex]\(a = 2\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -9\)[/tex].
We will use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, calculate the discriminant, [tex]\(\Delta\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 2^2 - 4(2)(-9) \][/tex]
[tex]\[ \Delta = 4 + 72 \][/tex]
[tex]\[ \Delta = 76 \][/tex]
Since the discriminant is positive, we will have two distinct real roots.
Next, we find the two solutions using the quadratic formula:
For [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x_1 = \frac{-2 + \sqrt{76}}{2(2)} \][/tex]
[tex]\[ x_1 = \frac{-2 + \sqrt{76}}{4} \][/tex]
[tex]\[ x_1 \approx \frac{-2 + 8.72}{4} \][/tex]
[tex]\[ x_1 \approx \frac{6.72}{4} \][/tex]
[tex]\[ x_1 \approx 1.68 \][/tex]
For [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x_2 = \frac{-2 - \sqrt{76}}{2(2)} \][/tex]
[tex]\[ x_2 = \frac{-2 - 8.72}{4} \][/tex]
[tex]\[ x_2 \approx \frac{-10.72}{4} \][/tex]
[tex]\[ x_2 \approx -2.68 \][/tex]
Thus, the solutions to the quadratic equation [tex]\(2x^2 + 2x - 9 = 0\)[/tex] are:
[tex]\[ x \approx 1.68, -2.68 \][/tex]
These results are rounded to the nearest hundredth.
