{"id":41387,"date":"2025-06-28T08:27:18","date_gmt":"2025-06-28T08:27:18","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=41387"},"modified":"2025-06-28T08:27:19","modified_gmt":"2025-06-28T08:27:19","slug":"find-the-square-roots-of-119-120i-algebraically","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-square-roots-of-119-120i-algebraically\/","title":{"rendered":"Find the square roots of 119 + 120i algebraically"},"content":{"rendered":"\n
Find the square roots of 119 + 120i algebraically. Let \u00e2\u02c6\u0161(119 + 120i) = p + qi be the square root of 119 + 120i. Then (\u00e2\u02c6\u0161(119 + 120i))^2 = 119 + 120i and (p + qi)^2 = 119 + 120i. a. Expand the left side of this equation. b. Equate the real and imaginary parts, and solve for p and q. c. What are the square roots of 119 + 120i<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
To find the square roots of the complex number 119+120i119 + 120i119+120i algebraically, let:119+120i=p+qi\\sqrt{119 + 120i} = p + qi119+120i\u200b=p+qi<\/p>\n\n\n\n
Then squaring both sides gives:(p+qi)2=119+120i(p + qi)^2 = 119 + 120i(p+qi)2=119+120i<\/p>\n\n\n\n
From the second equation:pq=60(Equation 1)pq = 60 \\quad \\text{(Equation 1)}pq=60(Equation 1)<\/p>\n\n\n\n
From the first equation:p2\u2212q2=119(Equation 2)p^2 – q^2 = 119 \\quad \\text{(Equation 2)}p2\u2212q2=119(Equation 2)<\/p>\n\n\n\n
Now use substitution. From Equation 1, q=60pq = \\frac{60}{p}q=p60\u200b. Plug into Equation 2:p2\u2212(60p)2=119p^2 – \\left(\\frac{60}{p}\\right)^2 = 119p2\u2212(p60\u200b)2=119p2\u22123600p2=119p^2 – \\frac{3600}{p^2} = 119p2\u2212p23600\u200b=119<\/p>\n\n\n\n
Multiply both sides by p2p^2p2 to eliminate the denominator:p4\u22123600=119p2p^4 – 3600 = 119p^2p4\u22123600=119p2p4\u2212119p2\u22123600=0p^4 – 119p^2 – 3600 = 0p4\u2212119p2\u22123600=0<\/p>\n\n\n\n
Solve using the quadratic formula:x=119\u00b11192+4\u22c536002x = \\frac{119 \\pm \\sqrt{119^2 + 4 \\cdot 3600}}{2}x=2119\u00b11192+4\u22c53600\u200b\u200bx=119\u00b114161+144002=119\u00b1285612x = \\frac{119 \\pm \\sqrt{14161 + 14400}}{2} = \\frac{119 \\pm \\sqrt{28561}}{2}x=2119\u00b114161+14400\u200b\u200b=2119\u00b128561\u200b\u200b28561=169\\sqrt{28561} = 16928561\u200b=169x=119\u00b11692\u21d2x=144 or x=\u221225x = \\frac{119 \\pm 169}{2} \\Rightarrow x = 144 \\text{ or } x = -25x=2119\u00b1169\u200b\u21d2x=144 or x=\u221225<\/p>\n\n\n\n
We discard x=\u221225x = -25x=\u221225 since x=p2x = p^2x=p2 must be nonnegative.<\/p>\n\n\n\n
So p2=144p^2 = 144p2=144, thus p=\u00b112p = \\pm12p=\u00b112<\/p>\n\n\n\n
Using Equation 1: pq=60pq = 60pq=60<\/p>\n\n\n\n
If p=12p = 12p=12, then q=6012=5q = \\frac{60}{12} = 5q=1260\u200b=5<\/p>\n\n\n\n
If p=\u221212p = -12p=\u221212, then q=60\u221212=\u22125q = \\frac{60}{-12} = -5q=\u22121260\u200b=\u22125<\/p>\n\n\n\n
\n\n\n\n
c. Final Answer<\/strong><\/h3>\n\n\n\n
The square roots of 119+120i119 + 120i119+120i are:12+5iand\u221212\u22125i\\boxed{12 + 5i \\quad \\text{and} \\quad -12 – 5i}12+5iand\u221212\u22125i\u200b<\/p>\n\n\n\n
\n\n\n\n
Explanation<\/strong><\/h3>\n\n\n\n
To find the square root of a complex number like 119+120i119 + 120i119+120i, we assume the root takes the form of p+qip + qip+qi, where both ppp and qqq are real numbers. Squaring this expression gives a new complex number, which we can compare to the original. Specifically:(p+qi)2=p2\u2212q2+2pqi(p + qi)^2 = p^2 – q^2 + 2pqi(p+qi)2=p2\u2212q2+2pqi<\/p>\n\n\n\n
By comparing this to 119+120i119 + 120i119+120i, we match the real part p2\u2212q2=119p^2 – q^2 = 119p2\u2212q2=119 and the imaginary part 2pq=1202pq = 1202pq=120. This gives us a system of equations. We isolate one variable using the second equation: pq=60pq = 60pq=60, so q=60pq = \\frac{60}{p}q=p60\u200b. Substituting into the first equation yields an equation with one variable:p2\u2212(60p)2=119p^2 – \\left(\\frac{60}{p}\\right)^2 = 119p2\u2212(p60\u200b)2=119<\/p>\n\n\n\n
To eliminate the fraction, we multiply through by p2p^2p2, leading to a quartic equation. After substitution x=p2x = p^2x=p2, the equation becomes a quadratic in xxx. Solving with the quadratic formula, we find the discriminant is a perfect square, which makes this problem solvable without approximation.<\/p>\n\n\n\n
The valid solution from the quadratic is x=144x = 144x=144, so p=\u00b112p = \\pm12p=\u00b112. Using pq=60pq = 60pq=60, we find the corresponding q=\u00b15q = \\pm5q=\u00b15, giving us two square roots: 12+5i12 + 5i12+5i and \u221212\u22125i-12 – 5i\u221212\u22125i. These are complex conjugates, as is typical for square roots of nonreal complex numbers.<\/p>\n\n\n\n
This algebraic method avoids converting to polar form and is useful when exact values are needed.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Find the square roots of 119 + 120i algebraically. Let \u00e2\u02c6\u0161(119 + 120i) = p + qi be the square root of 119 + 120i. Then (\u00e2\u02c6\u0161(119 + 120i))^2 = 119 + 120i and (p + qi)^2 = 119 + 120i. a. Expand the left side of this equation. b. Equate the real and imaginary […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-41387","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41387","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=41387"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41387\/revisions"}],"predecessor-version":[{"id":41395,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41387\/revisions\/41395"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/figure>\n","protected":false},"excerpt":{"rendered":"