{"id":35490,"date":"2025-06-24T11:24:11","date_gmt":"2025-06-24T11:24:11","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35490"},"modified":"2025-06-24T11:24:12","modified_gmt":"2025-06-24T11:24:12","slug":"solve-the-following-inequality","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/solve-the-following-inequality\/","title":{"rendered":"Solve the following inequality"},"content":{"rendered":"\n

Solve the following inequality. Write the solution in interval notation_ x2 3x – 4 > 0<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given the inequality:x2+3x\u22124>0x^2 + 3x – 4 > 0x2+3x\u22124>0<\/p>\n\n\n\n

Step 1: Factor the quadratic expression<\/h3>\n\n\n\n

We need to factor x2+3x\u22124x^2 + 3x – 4×2+3x\u22124. We look for two numbers that multiply to \u22124-4\u22124 and add up to 333. These numbers are 444 and \u22121-1\u22121.<\/p>\n\n\n\n

So,x2+3x\u22124=(x+4)(x\u22121)x^2 + 3x – 4 = (x + 4)(x – 1)x2+3x\u22124=(x+4)(x\u22121)<\/p>\n\n\n\n

Now the inequality becomes:(x+4)(x\u22121)>0(x + 4)(x – 1) > 0(x+4)(x\u22121)>0<\/p>\n\n\n\n

Step 2: Determine the critical points<\/h3>\n\n\n\n

Set each factor equal to zero:x+4=0\u21d2x=\u22124x\u22121=0\u21d2x=1x + 4 = 0 \\Rightarrow x = -4 \\\\ x – 1 = 0 \\Rightarrow x = 1x+4=0\u21d2x=\u22124x\u22121=0\u21d2x=1<\/p>\n\n\n\n

These are the critical points where the expression changes sign. They divide the number line into three intervals:<\/p>\n\n\n\n

    \n
  1. (\u2212\u221e,\u22124)(-\\infty, -4)(\u2212\u221e,\u22124)<\/li>\n\n\n\n
  2. (\u22124,1)(-4, 1)(\u22124,1)<\/li>\n\n\n\n
  3. (1,\u221e)(1, \\infty)(1,\u221e)<\/li>\n<\/ol>\n\n\n\n

    Step 3: Test each interval<\/h3>\n\n\n\n

    Pick a test point from each interval and plug it into (x+4)(x\u22121)(x + 4)(x – 1)(x+4)(x\u22121):<\/p>\n\n\n\n

      \n
    • In (\u2212\u221e,\u22124)(-\\infty, -4)(\u2212\u221e,\u22124), try x=\u22125x = -5x=\u22125:
      (x+4)(x\u22121)=(\u22125+4)(\u22125\u22121)=(\u22121)(\u22126)=6>0(x + 4)(x – 1) = (-5 + 4)(-5 – 1) = (-1)(-6) = 6 > 0(x+4)(x\u22121)=(\u22125+4)(\u22125\u22121)=(\u22121)(\u22126)=6>0<\/li>\n\n\n\n
    • In (\u22124,1)(-4, 1)(\u22124,1), try x=0x = 0x=0:
      (x+4)(x\u22121)=(0+4)(0\u22121)=(4)(\u22121)=\u22124<0(x + 4)(x – 1) = (0 + 4)(0 – 1) = (4)(-1) = -4 < 0(x+4)(x\u22121)=(0+4)(0\u22121)=(4)(\u22121)=\u22124<0<\/li>\n\n\n\n
    • In (1,\u221e)(1, \\infty)(1,\u221e), try x=2x = 2x=2:
      (x+4)(x\u22121)=(2+4)(2\u22121)=(6)(1)=6>0(x + 4)(x – 1) = (2 + 4)(2 – 1) = (6)(1) = 6 > 0(x+4)(x\u22121)=(2+4)(2\u22121)=(6)(1)=6>0<\/li>\n<\/ul>\n\n\n\n

      Step 4: Include correct intervals<\/h3>\n\n\n\n

      We want the product to be greater than zero<\/strong>, so we take the intervals where the product is positive. That is:<\/p>\n\n\n\n

        \n
      • (\u2212\u221e,\u22124)(-\\infty, -4)(\u2212\u221e,\u22124)<\/li>\n\n\n\n
      • (1,\u221e)(1, \\infty)(1,\u221e)<\/li>\n<\/ul>\n\n\n\n

        We do not<\/strong> include the endpoints \u22124-4\u22124 and 111 because the inequality is strictly greater than zero<\/strong> (not greater than or equal to zero).<\/p>\n\n\n\n

        Final Answer:<\/h3>\n\n\n\n

        (\u2212\u221e,\u22124)\u222a(1,\u221e)\\boxed{(-\\infty, -4) \\cup (1, \\infty)}(\u2212\u221e,\u22124)\u222a(1,\u221e)\u200b<\/p>\n\n\n\n

        Explanation<\/h3>\n\n\n\n

        Solving a quadratic inequality like x2+3x\u22124>0x^2 + 3x – 4 > 0x2+3x\u22124>0 involves a few important steps. First, we factor the quadratic expression. Factoring means rewriting the expression as a product of two binomials. In this case, the expression factors as (x+4)(x\u22121)(x + 4)(x – 1)(x+4)(x\u22121). This gives us a product of two linear expressions, and we now want to find when this product is greater than zero.<\/p>\n\n\n\n

        Next, we identify the critical points<\/strong>. These are the values of xxx that make the expression equal to zero. Setting each factor equal to zero gives x=\u22124x = -4x=\u22124 and x=1x = 1x=1. These values split the number line into three regions. Each region needs to be tested to determine whether the inequality holds there.<\/p>\n\n\n\n

        We pick one test point in each region and plug it into the factored form (x+4)(x\u22121)(x + 4)(x – 1)(x+4)(x\u22121). If the result is positive, the inequality is satisfied in that region. If it is negative, it is not. For x=\u22125x = -5x=\u22125 and x=2x = 2x=2, the product is positive. For x=0x = 0x=0, the product is negative.<\/p>\n\n\n\n

        Finally, we include only the intervals where the expression is positive. Since the inequality is strict (uses > and not \u2265), we do not include the boundary points x=\u22124x = -4x=\u22124 or x=1x = 1x=1.<\/p>\n\n\n\n

        Therefore, the solution in interval notation is:(\u2212\u221e,\u22124)\u222a(1,\u221e)(-\\infty, -4) \\cup (1, \\infty)(\u2212\u221e,\u22124)\u222a(1,\u221e)<\/p>\n\n\n\n

        This shows all values of xxx where the original inequality is true.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Solve the following inequality. Write the solution in interval notation_ x2 3x – 4 > 0 The Correct Answer and Explanation is: We are given the inequality:x2+3x\u22124>0x^2 + 3x – 4 > 0x2+3x\u22124>0 Step 1: Factor the quadratic expression We need to factor x2+3x\u22124x^2 + 3x – 4×2+3x\u22124. We look for two numbers that multiply […]<\/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-35490","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35490","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=35490"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35490\/revisions"}],"predecessor-version":[{"id":35492,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35490\/revisions\/35492"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}