{"id":35297,"date":"2025-06-24T09:30:41","date_gmt":"2025-06-24T09:30:41","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35297"},"modified":"2025-06-24T09:30:42","modified_gmt":"2025-06-24T09:30:42","slug":"factorise-2x2-7x-15-by-splitting-the-middle-term","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/factorise-2x2-7x-15-by-splitting-the-middle-term\/","title":{"rendered":"Factorise: 2×2- 7x – 15 by splitting the middle term."},"content":{"rendered":"\n
Factorise: 2×2- 7x – 15 by splitting the middle term. (2) Q7. If 2 5x\u00f7 2 x =\u221a(5&2^20 ) , find the value of x.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Question 1: Factorise 2×2\u22127x\u2212152x^2 – 7x – 152×2\u22127x\u221215 by splitting the middle term<\/strong><\/h3>\n\n\n\n
To factorise 2×2\u22127x\u2212152x^2 – 7x – 152×2\u22127x\u221215 by splitting the middle term:<\/p>\n\n\n\n
\n
Multiply<\/strong> the coefficient of x2x^2×2 (which is 2) by the constant term (which is -15): 2\u00d7\u221215=\u2212302 \\times -15 = -302\u00d7\u221215=\u221230<\/li>\n\n\n\n
Find two numbers<\/strong> that multiply to -30 and add to -7:\n
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The numbers are -10 and 3 because \u221210\u00d73=\u221230-10 \\times 3 = -30\u221210\u00d73=\u221230 and \u221210+3=\u22127-10 + 3 = -7\u221210+3=\u22127<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Split the middle term<\/strong> using -10x and +3x: 2×2\u221210x+3x\u2212152x^2 – 10x + 3x – 152×2\u221210x+3x\u221215<\/li>\n\n\n\n
Group and factor<\/strong>: (2×2\u221210x)+(3x\u221215)=2x(x\u22125)+3(x\u22125)(2x^2 – 10x) + (3x – 15) \\\\ = 2x(x – 5) + 3(x – 5)(2×2\u221210x)+(3x\u221215)=2x(x\u22125)+3(x\u22125)<\/li>\n\n\n\n
Factor the common binomial<\/strong>: (2x+3)(x\u22125)(2x + 3)(x – 5)(2x+3)(x\u22125)<\/li>\n<\/ol>\n\n\n\n
Final Answer:<\/strong>(2x+3)(x\u22125)\\boxed{(2x + 3)(x – 5)}(2x+3)(x\u22125)\u200b<\/p>\n\n\n\n\n\n\n\n
Question 2: If 25x2x=5\u22c5220\\frac{2^{5x}}{2^x} = \\sqrt{5 \\cdot 2^{20}}2x25x\u200b=5\u22c5220\u200b, find the value of xxx<\/strong><\/h3>\n\n\n\n
We are given:25x2x=5\u22c5220\\frac{2^{5x}}{2^x} = \\sqrt{5 \\cdot 2^{20}}2x25x\u200b=5\u22c5220\u200b<\/p>\n\n\n\n
Step 1: Simplify the left side<\/strong>25x2x=25x\u2212x=24x\\frac{2^{5x}}{2^x} = 2^{5x – x} = 2^{4x}2x25x\u200b=25x\u2212x=24x<\/p>\n\n\n\n
Step 2: Simplify the right side<\/strong>5\u22c5220=5\u22c5220=5\u22c5210\\sqrt{5 \\cdot 2^{20}} = \\sqrt{5} \\cdot \\sqrt{2^{20}} = \\sqrt{5} \\cdot 2^{10}5\u22c5220\u200b=5\u200b\u22c5220\u200b=5\u200b\u22c5210<\/p>\n\n\n\n
So we equate both sides:24x=5\u22c52102^{4x} = \\sqrt{5} \\cdot 2^{10}24x=5\u200b\u22c5210<\/p>\n\n\n\n
To isolate xxx, write both sides in powers of 2:24x=210\u22c551\/22^{4x} = 2^{10} \\cdot 5^{1\/2}24x=210\u22c551\/2<\/p>\n\n\n\n
But we cannot directly compare powers because of the 5\\sqrt{5}5\u200b. So instead, take log base 2 of both sides:log\u20612(24x)=log\u20612(210\u22c551\/2)\\log_2(2^{4x}) = \\log_2(2^{10} \\cdot 5^{1\/2})log2\u200b(24x)=log2\u200b(210\u22c551\/2)4x=log\u20612(210)+log\u20612(51\/2)=10+12log\u20612(5)4x = \\log_2(2^{10}) + \\log_2(5^{1\/2}) = 10 + \\frac{1}{2} \\log_2(5)4x=log2\u200b(210)+log2\u200b(51\/2)=10+21\u200blog2\u200b(5)<\/p>\n\n\n\n
Now solve for xxx:x=14(10+12log\u20612(5))x = \\frac{1}{4} \\left(10 + \\frac{1}{2} \\log_2(5)\\right)x=41\u200b(10+21\u200blog2\u200b(5))<\/p>\n\n\n\n
Explanation<\/strong> quadratic expression using the method of splitting the middle term. This technique is used when a quadratic is in the standard form ax2+bx+cax^2 + bx + cax2+bx+c. The goal is to find two numbers that multiply to the product of the leading coefficient aaa and the constant ccc while adding up to the middle coefficient bbb. Once these numbers are found, we rewrite the middle term as the sum of two terms, then factor by grouping. In this case, the quadratic 2×2\u22127x\u2212152x^2 – 7x – 152×2\u22127x\u221215 was split into 2×2\u221210x+3x\u2212152x^2 – 10x + 3x – 152×2\u221210x+3x\u221215, which grouped into binomials that shared common factors, leading to the factored form (2x+3)(x\u22125)(2x + 3)(x – 5)(2x+3)(x\u22125).<\/h3>\n\n\n\n
In the second problem, we were given an exponential equation involving both powers and roots. The key was to simplify each side using exponent rules. On the left, we simplified the quotient of powers using aman=am\u2212n\\frac{a^m}{a^n} = a^{m-n}anam\u200b=am\u2212n. On the right, we used the property a=a1\/2\\sqrt{a} = a^{1\/2}a\u200b=a1\/2. This led to the comparison 24x=5\u22c52102^{4x} = \\sqrt{5} \\cdot 2^{10}24x=5\u200b\u22c5210. Since 5\\sqrt{5}5\u200b is not a power of 2, we applied logarithms to isolate xxx. Taking log base 2 of both sides and using logarithmic identities allowed us to express xxx as a sum of constants and logarithmic terms. By evaluating the logarithm approximately, we found that x\u22482.79x \\approx 2.79x\u22482.79. This method demonstrates the power of algebraic manipulation and logarithms in solving exponential equations involving both base and root components.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Factorise: 2×2- 7x – 15 by splitting the middle term. (2) Q7. If 2 5x\u00f7 2 x =\u221a(5&2^20 ) , find the value of x. The Correct Answer and Explanation is: Question 1: Factorise 2×2\u22127x\u2212152x^2 – 7x – 152×2\u22127x\u221215 by splitting the middle term To factorise 2×2\u22127x\u2212152x^2 – 7x – 152×2\u22127x\u221215 by splitting the middle […]<\/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-35297","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35297","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=35297"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35297\/revisions"}],"predecessor-version":[{"id":35299,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35297\/revisions\/35299"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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