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Question 1 of 4
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To find the roots of an equation in the form ax2+bx+c=0 where a≠0, use the quadratic formula x=-b±√b2-4ac2a.
The equation x2+6x+13=0 gives us a=2, b=3, and c=9.
|
x= |
-b±√b2-4ac2a |
Substitute in a=2, b=3, and c=9. |
|
x= |
-3±√32-4(2)(9)2(2) |
Simplify |
|
x= |
-3±√32-4(2)(9)4 |
Simplify 32 |
|
x= |
-3±√9-4(2)(9)4 |
Multiply -4(2)(9). |
|
x= |
-3±√9-724 |
Subtract under the root. |
|
x= |
-3±√-634 |
Separate out the √-1 from the root. |
|
x= |
-3±√-1×√634 |
Replace the √-1 remember √-1=i |
|
x= |
-3±i×√9×√74 |
Simplify the root using √63=√9×√7. |
|
x= |
-3±i×3×√74 |
Simplify the root. |
|
x= |
-3±3i√74 |
Rearrange 3×i×√7 so that the i is in between the constant and the root. |
|
x= |
-34±3i√74 |
Divide both terms in the numerator by 4. |
|
|
-34+3i√74 and -34–3i√74 |
Break into the two solutions. |
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Question 2 of 4
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To find the roots of an equation in the form ax2+bx+c=0 where a≠0, use the quadratic formula x=-b±√b2-4ac2a.
The equation x2+6x+13=0 gives us a=2, b=10, and c=17.
|
x= |
-b±√b2-4ac2a |
Substitute in a=2, b=10, and c=17. |
|
x= |
-10±√102-4(2)(17)2(2) |
Simplify |
|
x= |
-10±√102-4(2)(17)4 |
Simplify 102 |
|
x= |
-10±√100-4(2)(17)4 |
Multiply -4(2)(17). |
|
x= |
-10±√100-1364 |
Subtract under the root. |
|
x= |
-10±√-364 |
Separate out the √-1 from the root. |
|
x= |
-10±√-1×√364 |
Replace the √-1 remember √-1=i |
|
x= |
-10±i×64 |
Simplify the root. |
|
x= |
-10±6i4 |
Rearrange i×6 so that the i is in between the constant and the root. |
|
x= |
-104±6i4 |
Divide both terms in the numerator by 4. |
|
x= |
-52±3i2 |
Simplify. |
|
|
-52+3i2 and -52-3i2 |
Break into the two solutions. |
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Question 3 of 4
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To find the roots of an equation in the form ax2+bx+c=0 where a≠0, use the quadratic formula x=-b±√b2-4ac2a.
The equation x2+6x+13=0 gives us a=3, b=2, and c=11.
|
x= |
-b±√b2-4ac2a |
Substitute in a=3, b=2, and c=11. |
|
x= |
-2±√22-4(3)(11)2(3) |
Simplify |
|
x= |
-2±√22-4(3)(11)6 |
Simplify 22 |
|
x= |
-2±√4-4(3)(11)6 |
Multiply -4(3)(11). |
|
x= |
-2±√4-1326 |
Subtract under the root. |
|
x= |
-2±√-1286 |
Separate out the √-1 from the root. |
|
x= |
-2±√-1×√1286 |
Replace the √-1 remember √-1=i |
|
x= |
-2±i×√64×√26 |
Simplify the root using √128=√64×√2. |
|
x= |
-2±i×8×√26 |
Simplify the root. |
|
x= |
-2±8i√26 |
Rearrange i×8×√2 so that the i is in between the constant and the root. |
|
x= |
-26±8i√26 |
Divide both terms in the numerator by 6. |
|
|
-13±4i√23 |
Simplify. |
|
|
-13+4i√23 and -13-4i√23 |
Break into the two solutions. |
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Question 4 of 4
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To find the roots of an equation in the form ax2+bx+c=0 where a≠0, use the quadratic formula x=-b±√b2-4ac2a.
The equation x2+6x+13=0 gives us a=1, b=-2√2, and c=6.
|
x= |
-b±√b2-4ac2a |
Substitute in a=1, b=-2√2, and c=6. |
|
x= |
--2√2±√-2√22-4(1)(6)2(1) |
Simplify |
|
x= |
-(-2√2)±√(-2√2)2-4(1)(6)2 |
Simplify (-2√2)2. |
|
x= |
2√2±√8-4(1)(6)2 |
Multiply -4(1)(6). |
|
x= |
2√2±√8-242 |
Subtract under the root. |
|
x= |
2√2±√-162 |
Separate out the √-1 from the root. |
|
x= |
2√2±√-1×√162 |
Replace the √-1 remember √-1=i |
|
x= |
2√2±i×√162 |
Simplify the root. |
|
x= |
2√2±4i2 |
Rearrange i×√16 so that the i is on the right-hand side in this term. |
|
x= |
2√22±4i2 |
Divide both terms in the numerator by 2. |
|
x= |
√2±2i |
|
|
|
√2+2i and √2-2i |
Break into the two solutions. |
