9:{"metadata":[["$","title","0",{"children":"iimkbmsat0025 - If one of the roots of the equation x^2 + ax + 3a = 0 is 1, then its other root is | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Looking at this prob<=m, I need to find the other root when one root is given as 1. Given Information: - Equation: x^2 + ax + 3a = 0 - One root is x = 1 - Need to find: the other root

Since x = 1 is a root, it must satisfy the equation. Let me substitute x = 1: 1^2 + a(1) + 3a = 0 1 + a + 3a = 0 1 + 4a = 0 4a = -1 a = -14

For any quadratic equation x^2 + bx + c = 0, if the roots are r_1 and r_2, then: - Sum of roots: r_1 + r_2 = -b In our equation x^2 + ax + 3a = 0: - b = a = -14 - Sum of roots = -a = -(-14) = 14 Since one root is 1, <=t the other root be r: 1 + r = 14 r = 14 - 1 = 14 - 44 = -34

Let me verify by substituting both roots back into the original equation with a = -14: The equation becomes: x^2 - 14x + 3(-14) = 0 Simplified: x^2 - 14x - 34 = 0 For x = 1: 1^2 - 14(1) - 34 = 1 - 14 - 34 = 1 - 1 = 0 ✓ For x = -34: (-34)^2 - 14(-34) - 34 = 916 + 316 - 1216 = 0 ✓ Answer: The other root is -34"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Algebra,Polynomials"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"iimkbmsat0025 - If one of the roots of the equation x^2 + ax + 3a = 0 is 1, then its other root is | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Looking at this prob<=m, I need to find the other root when one root is given as 1. Given Information: - Equation: x^2 + ax + 3a = 0 - One root is x = 1 - Need to find: the other root

Since x = 1 is a root, it must satisfy the equation. Let me substitute x = 1: 1^2 + a(1) + 3a = 0 1 + a + 3a = 0 1 + 4a = 0 4a = -1 a = -14

Let me verify by substituting both roots back into the original equation with a = -14: The equation becomes: x^2 - 14x + 3(-14) = 0 Simplified: x^2 - 14x - 34 = 0 For x = 1: 1^2 - 14(1) - 34 = 1 - 14 - 34 = 1 - 1 = 0 ✓ For x = -34: (-34)^2 - 14(-34) - 34 = 916 + 316 - 1216 = 0 ✓ Answer: The other root is -34"}],["$","meta","9",{"property":"og:url","content":"/iimk-bmsat-past-year-questions/2025/QA/7"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | JIPMAT Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"iimkbmsat0025 - If one of the roots of the equation x^2 + ax + 3a = 0 is 1, then its other root is | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Looking at this prob<=m, I need to find the other root when one root is given as 1. Given Information: - Equation: x^2 + ax + 3a = 0 - One root is x = 1 - Need to find: the other root

Since x = 1 is a root, it must satisfy the equation. Let me substitute x = 1: 1^2 + a(1) + 3a = 0 1 + a + 3a = 0 1 + 4a = 0 4a = -1 a = -14

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