{"id":2893,"date":"2023-10-03T16:00:18","date_gmt":"2023-10-03T16:00:18","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2893"},"modified":"2023-10-03T16:03:34","modified_gmt":"2023-10-03T16:03:34","slug":"week-55-ss2-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-55-ss2-first-term-mathematics-notes\/","title":{"rendered":"Week 55 &#8211; SS2 First Term Mathematics Notes"},"content":{"rendered":"<p>\u00a0<br \/>\n\u00a0<strong>WEEK FIVE<br \/>\n<\/strong><strong>QUADRATIC EQUATIONS<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Construction of Quadratic Equations from Sum and Product of Roots.\n<\/li>\n<li>Word Problem Leading to Quadratic Equations.\n<\/li>\n<\/ul>\n<p>\u00a0<strong>CONSTRUCTION OF QUADRATIC EQUATIONS FROM SUM AND PRODUCT OF ROOTS<br \/>\n<\/strong>We can find the sum and product of the roots directly from the coefficient in the equation. It is usual to call the roots of the equation \u03b1 and \u03b2  If the equation<br \/>\n\u00a0\u00a0\u00a0\u00a0ax<sup>2<\/sup> +bx + C = 0   \u2026\u2026\u2026\u2026\u2026.  I<br \/>\nhas the roots \u03b1 and \u03b2 then it is equivalent to the equation<br \/>\n\u00a0\u00a0\u00a0\u00a0(x \u2013 \u03b1  )( x \u2013 \u03b2 )  = 0<br \/>\nx<sup>2<\/sup> \u2013 \u03b2x \u2013  \u03b2x + \u03b1\u03b2  = 0 \u2026\u2026\u2026\u2026 2<br \/>\nDivide equation (i)by the coefficient  of x<sup>2<\/sup><br \/>\n\t\t ax<sup>2<\/sup>+ bx + C     = 0  \u2026\u2026\u2026\u2026   3<br \/>\naaa<br \/>\nComparing equations (2) and (3)<br \/>\nx<sup>2<\/sup> +  b x  +  C     = 0<br \/>\naa<br \/>\nx<sup>2<\/sup>  &#8211;  ( \u03b1 +\u03b2)x  + \u03b1\u03b2    = 0<br \/>\nthen<br \/>\n\u03b1+ \u03b2= -b<br \/>\na<br \/>\nand \u03b1\u03b2 = C<br \/>\na<br \/>\nFor any quadratic equation, ax<sup>2<\/sup> +bx + C = 0 with roots \u03b1 and \u03b2<br \/>\n\u03b1 + \u03b2  = -b<br \/>\na<br \/>\n\u03b1\u03b2 =  C<br \/>\na<br \/>\n<strong>Examples<br \/>\n<\/strong>1. If the roots of 3x<sup>2<\/sup> \u2013 4x \u2013 1 = 0 are \u03b1and \u03b2, find  \u03b1 + \u03b2 and \u03b1\u03b2<br \/>\n2. if \u03b1 and \u03b2are the roots of the equation<br \/>\n     3x<sup>2<\/sup> \u2013 4x \u2013 1 = 0 , find the value of<br \/>\n(a)     \u03b1     +  \u03b2<br \/>\n\u03b2          \u03b1<br \/>\n(b)   \u03b1   &#8211;   \u03b2<br \/>\n<strong>Solutions<br \/>\n<\/strong>1.  Since \u03b1 + \u03b2 =  -b<br \/>\na<br \/>\nComparing the given equation 3x<sup>2<\/sup> \u2013 4x \u2013 1= 0 with the general form<br \/>\nax<sup>2<\/sup> + bx + C = 0<br \/>\na = 3, b = -4, C = 1.<\/p>\n<p>\u00a0Then<br \/>\n\u03b1 + \u03b2 = -b  =  -(-4)<br \/>\na              3<br \/>\n        = + 4  =  +1 <sup>1<\/sup>\/<sub>3<\/sub><br \/>\n\t\t               3<br \/>\n\u03b1\u03b2 =C   =  -1    = -1<br \/>\na          3         3<br \/>\n2.a\u03b1 +  \u03b2    =  \u03b1<sup>2<\/sup> +\u03b2<sup>2<\/sup><br \/>\n\t\t\t\u03b2    \u03b1          \u03b1\u03b2<br \/>\n= (\u03b1 + \u03b2 )<sup>2<\/sup> &#8211; 2\u03b1\u03b2<br \/>\n\t\t\u03b1\u03b2<br \/>\nHere, comparing the given equation, with the general equation,<br \/>\na = 3, b = -4, C = &#8211; 1<br \/>\nfrom the solution of example 1 (since the given equation are the same ),<br \/>\n\u03b1 + \u03b2 = -b =  &#8211; (-4)   = +4<br \/>\n                           3           3<br \/>\n\u03b1\u03b2  = C   = &#8211; 1<br \/>\na         3<br \/>\nthen<br \/>\n\u03b1  +    \u03b2    =  ( \u03b1+ \u03b2 )<sup> 2<\/sup> \u2013 2 \u03b1\u03b2<br \/>\n\t\t\u03b2         \u03b1                  \u03b1\u03b2<br \/>\n=  (4\/3 )<sup>.2<\/sup> \u2013 2 ( &#8211; 1\/3 )<\/p>\n<ul>\n<li>1\/3\n<\/li>\n<\/ul>\n<p>=   16        \u00b1 2<br \/>\n   9           3<br \/>\n&#8211; 1<br \/>\n            3<br \/>\n=   16 + 6   \u00f7    -1\/3<br \/>\n         9<br \/>\n22     x   -3<br \/>\n         9          1<br \/>\n=   -22<br \/>\n       3<br \/>\nor  \u03b1 + \u03b2     = &#8211; 22   = &#8211; 7 1\/3<br \/>\n\u03b2        \u03b1           3<br \/>\nb)  Since<br \/>\n(\u03b1-\u03b2) <sup>2<\/sup> =\u03b1<sup>2<\/sup> + \u03b2<sup>&#8211; 2<\/sup> \u03b1 \u03b2<br \/>\nbut<br \/>\n\u03b1<sup>2<\/sup> + \u03b2<sup>2<\/sup> = ( \u03b1 + \u03b2)<sup>2<\/sup> -2 \u03b1 \u03b2<br \/>\n:.(\u03b1- \u03b2)<sup>2<\/sup> = ( \u03b1+ \u03b2 )<sup>2<\/sup> &#8211; 2\u03b1\u03b2 -2\u03b1\u03b2<br \/>\n(\u03b1 \u2013 \u03b2)<sup>2<\/sup> = (\u03b1 + \u03b2 )<sup>2<\/sup> &#8211; 4\u03b1 \u03b2<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F1.png\" alt=\"\"\/><br \/>\n\t\t:.( \u03b1 \u2013 \u03b2) = \u221a(\u03b1 + \u03b2 )<sup>2<\/sup> &#8211; 4\u03b1\u03b2<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F2.png\" alt=\"\"\/>( \u03b1 \u2013 \u03b2) =\u221a   (4\/3 )<sup>2<\/sup> \u2013 4 ( &#8211; <sup>1<\/sup>\/3 )<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F3.png\" alt=\"\"\/><br \/>\n\t\t= \u221a 16\/9  +<sup>4<\/sup>\/3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F4.png\" alt=\"\"\/><br \/>\n\t\t =    \u221a16 + 12<br \/>\n              9<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F5.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F6.png\" alt=\"\"\/><br \/>\n\t\t  =      \u221a28           =   \u221a28<br \/>\n            9                    3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F7.png\" alt=\"\"\/>    :. \u03b1 &#8211; \u03b2  =\u00a0\u00a0\u00a0\u00a0    \u221a28<br \/>\n                                 3<br \/>\n<strong>Evaluation<br \/>\n<\/strong>If \u03b1 and \u03b2 are the roots of the equation<br \/>\n2x<sup>2<\/sup> \u2013 11x + 5 = 0, find the value of<br \/>\na.   \u03b1 &#8211; \u03b2<br \/>\nb.     1          +        1<br \/>\n\t\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F8.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week55SS2F9.png\" alt=\"\"\/>       \u03b1 + 1             \u03b2+ 1<strong><br \/>\n\t\t\t<\/strong>c. \u03b1<sup>2<\/sup>+  \u03b2<sup>2<\/sup><\/p>\n<p>\u00a0<strong>WORD PROBLEM LEADING TO QUADRATIC EQUATIONS<br \/>\n<\/strong><strong>Examples<br \/>\n<\/strong>1. Find two numbers whose difference is 5 and whose product is 266.<br \/>\n<strong>Solution<br \/>\n<\/strong>Let the smaller number be x.<br \/>\nThen the smaller number be x+5.<br \/>\nTheir product is x(x+5) .<\/p>\n<p>\u00a0Hence,<br \/>\nx(x+5) = 266<br \/>\nx<sup>2<\/sup>+5x-  266 = 0<br \/>\n(x-14)(x+19)=0<br \/>\n x=14 or x= -19<\/p>\n<p>\u00a0The other number is 14+5 or  -19+5 i.e 19 or -14<br \/>\n:. The two numbers are 14 and 19 or -14 and -14.<\/p>\n<p>\u00a02. Tina is 3 times older than her daughter. In four years time, the product of their ages will be 1536. How old are they now?<\/p>\n<p>\u00a0<strong>Solution<br \/>\n<\/strong>Let the daughter&#8217;s age be x.<br \/>\nTina&#8217;s age = 3x<br \/>\nIn four years&#8217; time,<br \/>\nDaughter&#8217;s age = (x+4)years<br \/>\nTina&#8217;s age = (3x+4)years<br \/>\nThe product of their ages :<br \/>\n (x+4)(3x+4)= 1536<br \/>\n3x<sup>2<\/sup>+ 16x \u2013 1520 = 0<br \/>\n(x-20)(3x+76) = =0<br \/>\n x=20 or x=-25.3<\/p>\n<p>\u00a0Since age cannot be negative, x=20years.<br \/>\n:. Daughter&#8217;s age = 20years.<br \/>\n Tina&#8217;s age = 20&#215;3=60years.   <\/p>\n<p>\u00a0<strong>Evaluation<br \/>\n<\/strong>1. Think of a number, square it, add 2 times the original number. The result is 80. Find the number.<br \/>\n2. The area of a square is 144cm<sup>2<\/sup> and one of its sides is (x+2)cm. Find x and then the side of the square.<br \/>\n3. Find two consecutive odd numbers whose product is 224.<\/p>\n<p>\u00a0<br \/>\n\u00a0<strong>GENERAL EVALUATION\/REVISION   QUESTIONS<br \/>\n<\/strong>1. The area of a rectangle is 60cm<sup>2<\/sup>. The length is 11cm more than the width. Find the width.<br \/>\n2. A man is 37years old and his child is 8. How many years ago was the product of their ages 96?<br \/>\n3. If \u03b1 and \u03b2 are the roots of the equation 2x<sup>2<\/sup> \u2013 9x+4=0, find<br \/>\na) \u03b1 + \u03b2      (b) \u03b1\u03b2       (c) \u03b1 \u2013 \u03b2   (d)  \u03b1\u03b2\/ \u03b1 + \u03b2      <\/p>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong>If \u03b1 and \u03b2 are the roots of the equation 2x<sup>2<\/sup> + 9x+9=0:<br \/>\n1. Find the product of their roots.     A. 4  B. 4.5   C. 5.5    B. -4.5<br \/>\n2. Find the sum of their roots.                 A. 4  B. 4.5   C. 5.5    B. -4.5<br \/>\n3. Find \u03b1<sup>2<\/sup>+\u03b2<sup>2<\/sup>                                          A. 11.5    B. -11.25   C. 11.25   D. -11.5<br \/>\n5. Find  \u03b1\u03b2\/ \u03b1 + \u03b2                                    A. 1  B.-1    C.  1.5     D. 4.5<\/p>\n<p>\u00a0<strong>THEORY<br \/>\n<\/strong>1. The base of a triangle is 3cm longer than its corresponding height. If the area is 44cm<sup>2<\/sup>, find the length of its base.<br \/>\n2. Find the equation in the form ax<sup>2<\/sup>+bx+c=0 whose sum and products of roots are respectively:<br \/>\na)  3,4    (b)  -7\/3  , 0    (c) 1.2,0.8<\/p>\n<p>\u00a0<strong>Reading Assignment<br \/>\n<\/strong>Essential Mathematics for SSS2, pages 50-54, exercise 4.6 and<br \/>\n<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0 \u00a0WEEK FIVE QUADRATIC EQUATIONS CONTENT Construction of Quadratic Equations from Sum and Product of&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,233],"tags":[],"class_list":["post-2893","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss2-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2893","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2893"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2893\/revisions"}],"predecessor-version":[{"id":2894,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2893\/revisions\/2894"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2893"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2893"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2893"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}