9:{"metadata":[["$","title","0",{"children":"iimkbmsat0025 - The zeroes of ax² + bx + c = 0 are in the ratio 1:2 (assume the coefficients to be non-zero). Which of these is one of the zeroes? | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Since the zeroes are in ratio 1:2, <=t them be k and 2k where k is some number.

For any quadratic ax^2 + bx + c = 0: - Sum of zeroes = -ba - Product of zeroes = ca

Sum of zeroes: k + 2k = 3k = -ba Therefore: k = -b3a

Product of zeroes: k 2k = 2k^2 = ca Therefore: k^2 = c2a

Substituting k = -b3a into k^2 = c2a: (-b3a)^2 = c2a b^29a^2 = c2a Cross-multiplying: 2ab^2 = 9a^2c Since a 0: 2b^2 = 9ac

The two zeroes are: - First zero: k = -b3a - Second zero: 2k = -2b3a Therefore, one of the zeroes is -b3a or -2b3a."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$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"}],["$","meta","7",{"property":"og:title","content":"iimkbmsat0025 - The zeroes of ax² + bx + c = 0 are in the ratio 1:2 (assume the coefficients to be non-zero). Which of these is one of the zeroes? | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Since the zeroes are in ratio 1:2, <=t them be k and 2k where k is some number.

For any quadratic ax^2 + bx + c = 0: - Sum of zeroes = -ba - Product of zeroes = ca

Sum of zeroes: k + 2k = 3k = -ba Therefore: k = -b3a

Product of zeroes: k 2k = 2k^2 = ca Therefore: k^2 = c2a

Substituting k = -b3a into k^2 = c2a: (-b3a)^2 = c2a b^29a^2 = c2a Cross-multiplying: 2ab^2 = 9a^2c Since a 0: 2b^2 = 9ac

The two zeroes are: - First zero: k = -b3a - Second zero: 2k = -2b3a Therefore, one of the zeroes is -b3a or -2b3a."}],["$","meta","9",{"property":"og:url","content":"/iimk-bmsat-past-year-questions/2025/QA/30"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"iimkbmsat0025 - The zeroes of ax² + bx + c = 0 are in the ratio 1:2 (assume the coefficients to be non-zero). Which of these is one of the zeroes? | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Since the zeroes are in ratio 1:2, <=t them be k and 2k where k is some number.

For any quadratic ax^2 + bx + c = 0: - Sum of zeroes = -ba - Product of zeroes = ca

Sum of zeroes: k + 2k = 3k = -ba Therefore: k = -b3a

Product of zeroes: k 2k = 2k^2 = ca Therefore: k^2 = c2a

Substituting k = -b3a into k^2 = c2a: (-b3a)^2 = c2a b^29a^2 = c2a Cross-multiplying: 2ab^2 = 9a^2c Since a 0: 2b^2 = 9ac

The two zeroes are: - First zero: k = -b3a - Second zero: 2k = -2b3a Therefore, one of the zeroes is -b3a or -2b3a."}],["$","meta","17",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","link","18",{"rel":"shortcut icon","href":"https://www.afterboards.in/favicon.ico","type":"image/x-icon","sizes":"48x48"}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f"],"error":null,"digest":"$undefined"}

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