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In Mathematics / High School | 2025-07-03

Solve the two equations in the table by factoring. Then enter those factors and the solutions in the table.

Equation

[tex]k^2+10=0[/tex]

[tex]4 x^2+29=0[/tex]

Solutions

Asked by fredo1234551

Answer (2)

The solutions for the first equation k 2 + 10 = 0 are k = ± i 10 ​ with the factors being ( k − i 10 ​ ) ( k + i 10 ​ ) = 0 . For the second equation 4 x 2 + 29 = 0 , the solutions are x = ± i 2 29 ​ ​ with the factors ( 2 x − i 29 ​ ) ( 2 x + i 29 ​ ) = 0 .
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Answered by Anonymous | 2025-07-04

Solve k 2 + 10 = 0 to find k = ± i 10 ​ .
Factor the equation as ( k − i 10 ​ ) ( k + i 10 ​ ) = 0 .
Solve 4 x 2 + 29 = 0 to find x = ± i 2 29 ​ ​ .
Factor the equation as ( 2 x − i 29 ​ ) ( 2 x + i 29 ​ ) = 0 .
The solutions are k = ± i 10 ​ and x = ± i 2 29 ​ ​ .

Explanation

Understanding the Problem We are given two equations: k 2 + 10 = 0 and 4 x 2 + 29 = 0 . We need to solve these equations by factoring and find their solutions.

Solving the First Equation Let's solve the first equation, k 2 + 10 = 0 . We can rewrite this as k 2 = − 10 . Taking the square root of both sides, we get k = ± − 10 ​ . Since we have a negative number under the square root, we can express the solution in terms of the imaginary unit i , where i = − 1 ​ . Thus, k = ± i 10 ​ . The factors are ( k − i 10 ​ ) and ( k + i 10 ​ ) .

Solving the Second Equation Now let's solve the second equation, 4 x 2 + 29 = 0 . We can rewrite this as 4 x 2 = − 29 . Dividing both sides by 4, we get x 2 = − 4 29 ​ . Taking the square root of both sides, we get x = ± − 4 29 ​ ​ . Again, we have a negative number under the square root, so we can express the solution in terms of the imaginary unit i . Thus, x = ± i 2 29 ​ ​ . The factors are ( 2 x − i 29 ​ ) and ( 2 x + i 29 ​ ) .

Final Solutions Therefore, the solutions for the first equation are k = i 10 ​ and k = − i 10 ​ , and the solutions for the second equation are x = i 2 29 ​ ​ and x = − i 2 29 ​ ​ .


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