{"id":35494,"date":"2025-06-24T11:28:29","date_gmt":"2025-06-24T11:28:29","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35494"},"modified":"2025-06-24T11:28:31","modified_gmt":"2025-06-24T11:28:31","slug":"solve-the-following-inequality-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/solve-the-following-inequality-2\/","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 asked to solve the inequality:x2+3x\u22124>0x^2 + 3x – 4 > 0x2+3x\u22124>0<\/p>\n\n\n\n

\n\n\n\n

Step 1: Solve the Related Equation<\/strong><\/h3>\n\n\n\n

To solve the inequality, first find the roots of the corresponding equation:x2+3x\u22124=0x^2 + 3x – 4 = 0x2+3x\u22124=0<\/p>\n\n\n\n

Use the quadratic formula:x=\u2212b\u00b1b2\u22124ac2ax = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}x=2a\u2212b\u00b1b2\u22124ac\u200b\u200b<\/p>\n\n\n\n

Here, a=1a = 1a=1, b=3b = 3b=3, and c=\u22124c = -4c=\u22124. Plugging into the formula:x=\u22123\u00b132\u22124(1)(\u22124)2(1)=\u22123\u00b19+162=\u22123\u00b1252x = \\frac{-3 \\pm \\sqrt{3^2 – 4(1)(-4)}}{2(1)} = \\frac{-3 \\pm \\sqrt{9 + 16}}{2} = \\frac{-3 \\pm \\sqrt{25}}{2}x=2(1)\u22123\u00b132\u22124(1)(\u22124)\u200b\u200b=2\u22123\u00b19+16\u200b\u200b=2\u22123\u00b125\u200b\u200bx=\u22123\u00b152x = \\frac{-3 \\pm 5}{2}x=2\u22123\u00b15\u200b<\/p>\n\n\n\n

This gives two real roots:x=\u22123+52=1andx=\u22123\u221252=\u22124x = \\frac{-3 + 5}{2} = 1 \\quad \\text{and} \\quad x = \\frac{-3 – 5}{2} = -4x=2\u22123+5\u200b=1andx=2\u22123\u22125\u200b=\u22124<\/p>\n\n\n\n

\n\n\n\n

Step 2: Determine the Sign of the Expression in Each Interval<\/strong><\/h3>\n\n\n\n

The roots 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

    Choose a test point in each interval to determine the sign of the expression x2+3x\u22124x^2 + 3x – 4×2+3x\u22124.<\/p>\n\n\n\n

      \n
    • Interval 1: x=\u22125x = -5x=\u22125<\/strong> (\u22125)2+3(\u22125)\u22124=25\u221215\u22124=6>0(-5)^2 + 3(-5) – 4 = 25 – 15 – 4 = 6 > 0(\u22125)2+3(\u22125)\u22124=25\u221215\u22124=6>0<\/li>\n\n\n\n
    • Interval 2: x=0x = 0x=0<\/strong> 02+3(0)\u22124=\u22124<00^2 + 3(0) – 4 = -4 < 002+3(0)\u22124=\u22124<0<\/li>\n\n\n\n
    • Interval 3: x=2x = 2x=2<\/strong> 22+3(2)\u22124=4+6\u22124=6>02^2 + 3(2) – 4 = 4 + 6 – 4 = 6 > 022+3(2)\u22124=4+6\u22124=6>0<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 3: Interpret the Results<\/strong><\/h3>\n\n\n\n

      We want the expression to be greater than zero<\/strong>, so we select the intervals where the expression is positive<\/strong>. From our testing:<\/p>\n\n\n\n

        \n
      • Positive in (\u2212\u221e,\u22124)(-\\infty, -4)(\u2212\u221e,\u22124) and (1,\u221e)(1, \\infty)(1,\u221e)<\/li>\n\n\n\n
      • Not included at x=\u22124x = -4x=\u22124 and x=1x = 1x=1 because the inequality is strictly greater than zero<\/strong> and not inclusive<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Final Answer in Interval Notation<\/strong><\/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

        \n\n\n\n

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

        To solve the inequality x2+3x\u22124>0x^2 + 3x – 4 > 0x2+3x\u22124>0, we begin by analyzing the related quadratic equation. The key is to understand where the quadratic expression is positive. First, we find the roots of the equation x2+3x\u22124=0x^2 + 3x – 4 = 0x2+3x\u22124=0 using the quadratic formula. After simplifying, we find the roots are x=\u22124x = -4x=\u22124 and x=1x = 1x=1. These roots divide the number line into three intervals: less than -4, between -4 and 1, and greater than 1.<\/p>\n\n\n\n

        Next, we pick a test value in each interval to check whether the expression is positive or negative in that section. For example, choosing x=\u22125x = -5x=\u22125 in the interval less than -4, we substitute into the original expression and find that it results in a positive value. This indicates the expression is greater than zero on that interval. Doing the same for the other two intervals, we find that the expression is negative between -4 and 1, and again positive when x>1x > 1x>1.<\/p>\n\n\n\n

        Since the original inequality asks for where the expression is strictly greater than zero, we exclude the values x=\u22124x = -4x=\u22124 and x=1x = 1x=1, because the expression equals zero at those points, not greater than zero. Therefore, the final solution includes only the intervals where the expression is strictly positive: from negative infinity to -4, and from 1 to positive infinity.<\/p>\n\n\n\n

        This approach works for any quadratic inequality: find the roots, use test intervals to check signs, and determine which regions satisfy the inequality. Express the final answer in interval notation, excluding boundary points for strict inequalities.<\/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 asked to solve the inequality:x2+3x\u22124>0x^2 + 3x – 4 > 0x2+3x\u22124>0 Step 1: Solve the Related Equation To solve the inequality, first find the roots of the corresponding equation:x2+3x\u22124=0x^2 + 3x […]<\/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-35494","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35494","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=35494"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35494\/revisions"}],"predecessor-version":[{"id":35496,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35494\/revisions\/35496"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35494"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35494"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35494"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}