Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To determine which of the given choices is a solution to the equation [tex]\((x-2)(x+10)=13\)[/tex], follow these steps:
1. Set up the equation:
[tex]\[(x-2)(x+10)= 13\][/tex]
2. Expand the left-hand side of the equation:
When expanding [tex]\((x-2)(x+10)\)[/tex], use the distributive property (also known as the FOIL method for binomials):
[tex]\[ (x-2)(x+10) = x^2 + 10x - 2x - 20 = x^2 + 8x - 20 \][/tex]
So, the equation becomes:
[tex]\[ x^2 + 8x - 20 = 13 \][/tex]
3. Move all terms to one side to form a standard quadratic equation:
Subtract 13 from both sides:
[tex]\[ x^2 + 8x - 20 - 13 = 0 \][/tex]
Simplify the equation:
[tex]\[ x^2 + 8x - 33 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(x^2 + 8x - 33 = 0\)[/tex]:
This can be done by factoring, completing the square, or using the quadratic formula. We can use the quadratic formula in this case, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = -33\)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-33)}}{2 \cdot 1} \][/tex]
Simplify under the square root:
[tex]\[ x = \frac{-8 \pm \sqrt{64 + 132}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{196}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm 14}{2} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{-8 + 14}{2} = \frac{6}{2} = 3 \][/tex]
and
[tex]\[ x = \frac{-8 - 14}{2} = \frac{-22}{2} = -11 \][/tex]
5. Check which of the provided choices match the solutions:
The possible choices are:
[tex]\[ x = 3, \quad x = 8, \quad x = 10, \quad x = 11 \][/tex]
From our solutions, we have [tex]\(x = 3\)[/tex] and [tex]\(x = -11\)[/tex]. Therefore, among the provided choices, the correct solution is:
[tex]\[ \boxed{3} \][/tex]
1. Set up the equation:
[tex]\[(x-2)(x+10)= 13\][/tex]
2. Expand the left-hand side of the equation:
When expanding [tex]\((x-2)(x+10)\)[/tex], use the distributive property (also known as the FOIL method for binomials):
[tex]\[ (x-2)(x+10) = x^2 + 10x - 2x - 20 = x^2 + 8x - 20 \][/tex]
So, the equation becomes:
[tex]\[ x^2 + 8x - 20 = 13 \][/tex]
3. Move all terms to one side to form a standard quadratic equation:
Subtract 13 from both sides:
[tex]\[ x^2 + 8x - 20 - 13 = 0 \][/tex]
Simplify the equation:
[tex]\[ x^2 + 8x - 33 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(x^2 + 8x - 33 = 0\)[/tex]:
This can be done by factoring, completing the square, or using the quadratic formula. We can use the quadratic formula in this case, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = -33\)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-33)}}{2 \cdot 1} \][/tex]
Simplify under the square root:
[tex]\[ x = \frac{-8 \pm \sqrt{64 + 132}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{196}}{2} \][/tex]
[tex]\[ x = \frac{-8 \pm 14}{2} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{-8 + 14}{2} = \frac{6}{2} = 3 \][/tex]
and
[tex]\[ x = \frac{-8 - 14}{2} = \frac{-22}{2} = -11 \][/tex]
5. Check which of the provided choices match the solutions:
The possible choices are:
[tex]\[ x = 3, \quad x = 8, \quad x = 10, \quad x = 11 \][/tex]
From our solutions, we have [tex]\(x = 3\)[/tex] and [tex]\(x = -11\)[/tex]. Therefore, among the provided choices, the correct solution is:
[tex]\[ \boxed{3} \][/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.