{"id":311,"date":"2017-07-08T00:32:55","date_gmt":"2017-07-07T19:02:55","guid":{"rendered":"http:\/\/ibelitetutor.com\/blog\/?p=311"},"modified":"2023-08-14T12:49:51","modified_gmt":"2023-08-14T07:19:51","slug":"quadratic-functions","status":"publish","type":"post","link":"http:\/\/ibelitetutor.com\/blog\/quadratic-functions\/","title":{"rendered":"Quadratic equations, Quadratic Functions and quadratic Formula"},"content":{"rendered":"

Quadratic equations, Quadratic Functions<\/span><\/h2>\n

Many IB Maths Tutors<\/a><\/strong> consider quadratic equations<\/strong> as a very important topic of maths. There are the following ways to solve a quadratic equation<\/p>\n

\u25ba Factorization method<\/strong><\/span><\/p>\n

\u25bacomplete square method<\/strong>\u00a0<\/span><\/p>\n

\u25ba graphical method<\/strong><\/span><\/p>\n

\u25ba Quadratic formula method<\/strong><\/span><\/p>\n

Quadratic equations, Quadratic Functions, and Quadratic Formula<\/strong><\/span><\/h2>\n

The quadratic formula is the strongest method to solve a quadratic equation. In this article, I will use a few steps to prove\u00a0the quadratic\u00a0formula.<\/strong><\/p>\n

Given equation:\u00a0ax\u00b2+bx+c=0<\/p>\n

Step-1:<\/strong> transfer constant term to the right side<\/p>\n

ax\u00b2+bx=-c<\/p>\n

Step-2:<\/strong> divide both sides by coefficient of x\u00b2<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 x\u00b2+bx\/a=-c\/a<\/div>\n
<\/div>\n
Step-3:<\/strong> write (coefficient of x\/2)\u00b2 \u00a0 \u00a0 that is (b\/2a)\u00b2=b\u00b2\/4a\u00b2<\/div>\n
<\/div>\n
<\/div>\n
Step-4:<\/strong> Add this value to both sides<\/div>\n
<\/div>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 x\u00b2+bx\/a+b\u00b2\/4a\u00b2 =-c\/a\u00b2+b\u00b2\/4a\u00b2<\/div>\n
<\/div>\n
<\/div>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (x+b\/2a)\u00b2=b\u00b2-4ac\/4a\u00b2<\/div>\n
<\/div>\n
now, take square root on both sides<\/div>\n
\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 x+b\/2=\u00b1\u221ab\u00b2-4ac\/a\u00b2<\/p>\n

<\/div>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0x=-(b\/2a)\u00b1\u221ab\u00b2-4ac\/2a<\/div>\n
<\/div>\n
\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>\"x <\/p>\n

This formula is known as the quadratic formula. We have used a simple way to\u00a0prove quadratic\u00a0formula<\/strong> we can put values\u00a0of a, b and c \u00a0from any equation and find the value of x (the variable) by directly using this formula.<\/span><\/p>\n<\/div>\n<\/div>\n

<\/div>\n
IB Mathematics tutors can also explain the concept of conjugate roots with the help of the quadratic formula. In a quadratic equation,<\/div>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ax\u00b2+bx+c=0<\/div>\n
if a, b and c are all rational numbers and one root of the quadratic equation is a+\u221ab<\/strong><\/span> then the second root will automatically become a-\u221ab<\/span>. that can be understood easily as we use one +ve<\/span> and one -ve<\/span> sign in the quadratic formula.
\nThese types of roots are called Conjugate Roots<\/strong>.<\/div>\n
<\/div>\n
If \u00a0a & b \u00a0are \u00a0the \u00a0roots \u00a0of \u00a0the \u00a0quadratic \u00a0equation \u00a0ax\u00b2 + bx + c = 0, \u00a0then;<\/div>\n
<\/div>\n
(i) \u00a0\"\\alpha\u00a0 (ii) \u00a0\"\\alpha\u00a0 \u00a0 \u00a0(iii) \u00a0\"\\alpha<\/div>\n
<\/div>\n
Nature \u00a0Of \u00a0Roots:<\/strong><\/div>\n
<\/div>\n
(a) Consider the quadratic equation ax\u00b2 + bx + c = 0 \u00a0where a, b, c\u00a0\"\u00a0R &\u00a0\"a\u00a0then<\/div>\n
<\/div>\n
(i) D > 0 \u00a0\"\u00a0roots \u00a0are \u00a0real & distinct \u00a0(unequal).<\/div>\n
<\/div>\n
(ii) D = 0\u00a0\"\u00a0roots \u00a0are \u00a0real & coincident \u00a0(equal).<\/div>\n
<\/div>\n
(iii) D < 0\"\u00a0roots \u00a0are \u00a0imaginary<\/div>\n
<\/div>\n
(B) Consider the quadratic equation\u00a0ax<\/span>2<\/sup>+ bx + c = 0 where a, b, c\u00a0\"\u00a0Q &\u00a0\"a\u00a0then<\/span><\/div>\n
\u00a0If \u00a0D > 0 \u00a0& \u00a0is a perfect \u00a0square , then \u00a0roots \u00a0are \u00a0rational & unequal. <\/span><\/div>\n
<\/div>\n
A quadratic \u00a0equation \u00a0whose \u00a0roots \u00a0are \u00a0a & b \u00a0is \u00a0(x – a)(x – b) = 0 \u00a0i.e. \u00a0 x2<\/sup> – (a + b) x + a b = 0 i.e.<\/span><\/div>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0x2<\/sup> – (sum of \u00a0roots) x + \u00a0product \u00a0of \u00a0roots = 0<\/strong><\/span><\/div>\n
<\/div>\n
Consider \u00a0the \u00a0quadratic \u00a0expression , y = ax\u00b2 + bx + c \u00a0, a, b, c\u00a0\"\u00a0R &\u00a0\"a\u00a0 then <\/span><\/div>\n
<\/div>\n
(i) The graph between x, y \u00a0is always a \u00a0parabola. \u00a0If a > 0 \u00a0then the shape of the parabola is concave upwards & \u00a0if a < 0 \u00a0then the shape of the parabola is concave downwards.<\/span><\/div>\n
<\/div>\n
Common \u00a0Roots \u00a0Of \u00a02 \u00a0Quadratic \u00a0Equations \u00a0[Only \u00a0One \u00a0Common \u00a0Root]-<\/strong>\u00a0 \u00a0Let \u00a0\"\\alpha
\nbe \u00a0the \u00a0common \u00a0root \u00a0of \u00a0ax\u00b2 + bx + c = 0 \u00a0& \u00a0a\u2019x<\/span>2<\/sup> + b\u2019x + c\u2019 = 0<\/span>
\nTherefore \u00a0 \u00a0\"\\;a{\\alpha<\/div>\n
\u00a0If we solve above pair by cramer’s rule we get<\/div>\n
\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\"\\frac{{{\\alpha<\/p>\n

<\/p>\n<\/div>\n

<\/div>\n
This will give us \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\"\\alpha<\/div>\n
<\/div>\n
\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\"{(ac'<\/p>\n

<\/p>\n

Every pair of the quadratic equation whose coefficients fulfills the above condition will have one root in common.<\/p>\n<\/div>\n

The condition that a quadratic function- \u00a0<\/strong><\/div>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0f(x , y) = ax\u00b2 + 2 hxy + by\u00b2 + 2 gx + 2 fy + c \u00a0may be \u00a0resolved \u00a0into \u00a0two \u00a0linear \u00a0factors \u00a0is \u00a0that \u00a0 \u00a0\u00a0\"abc{\\rm{\u00a0or<\/div>\n
<\/div>\n
\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\"\\left|<\/div>\n
Reducible Quadratic Equations-<\/strong>These are the equations which are not quadratic in their initial condition but after some calculations, we can reduce them into quadratic equations<\/div>\n
<\/div>\n
(i) If the power of the second term is exactly half to the power of the first term and the third term is a constant, these types of equations can be reduced to quadratic equations.
\ni.e.,\u00a0\"x\u00a0,\u00a0\"{x^6},\u00a0\"{e^{2x}}\u00a0,\"{a^{2x}}\"{\\left( \u00a0all these equations can easily be reduced into quadratic equations by applying the method of substitution.<\/div>\n
Example-<\/strong>Solve this equation and find x<\/div>\n
\n

\u00a0\"{\\left(\u00a0<\/strong>Ans:<\/p>\n

let y= \u00a0\"\\left(\u00a0then given equation will become<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\"{y^2}\u00a0 it’s a simple quadratic equation we can be easily factorized it and solve \u00a0so y=–3,2<\/p>\n

so\u00a0\u00a0\"\\left(=-3<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\"\\frac{{4{x^2} <\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\"\\begin{array}{l}<\/p>\n

<\/p>\n

the final equation can be solved using Quadratic formula and the same process can be repeated for \u00a0y=2<\/p>\n

<\/p>\n

(ii) If a variable is added with it is own reciprocal, then we get a quadratic equation i.e,\"x\"\\frac{{2x\"{e^{2x}}\"{a^{2x}}\u00a0all these equations can be reduced into quadratic by replacing one term by any other variable.<\/p>\n

Standard Form of a Quadratic Function<\/strong>-A quadratic function y=ax2<\/sup>+ bx + c can be<\/p>\n

reduced into standard form \u00a0 \u00a0\"y\u00a0 by the method of completing the square. If we<\/p>\n

draw the graph of this function we shall get a parabola with vertex (h,k). The parabola will be upward for a>0 and downward for a<0<\/p>\n

Maximum and Minimum value of a quadratic function- <\/strong>If the function is in the form<\/p>\n

\"y\u00a0Then ‘h’ is the input value of the function while ‘k’ is its <\/span>output.<\/p>\n

(i) If a>0 (in case of the upward parabola) the minimum value of f is f(h)=k<\/p>\n<\/div>\n

(ii) If a<0 (in case of a downward parabola)the maximum value of f is f(h)=k<\/div>\n
<\/div>\n
If our function is in the form of \u00a0y=ax\u00b2 + bx + c then vertex of the parabola \"V<\/div>\n
The line passing through vertex and parallel to the y-axis is called the axis of symmetry.<\/div>\n
The parabolic graph of a quadratic function is symmetrical about axis of symmetry.<\/div>\n
\n

\"\u00a0f(x) has a minimum value at vertex if a>0 and\u00a0\"{f_{\\min\u00a0\u00a0at \u00a0\u00a0\"x<\/p>\n

\"\u00a0\u00a0f(x) has a maximum value at vertex if a<0 and \u00a0\u00a0\"{f_{\\min\u00a0\u00a0at \u00a0\u00a0\"x<\/p>\n<\/div>\n

In the next post about quadratics, I shall discuss discriminant, nature of roots, relationships between the roots. In the meantime, you can download the pdf and solve practice questions<\/p>\n

\u00a0\"ib<\/div>\n
\n
\n

<\/p>

    <\/ul><\/div>\n
    \n
    \n<\/fieldset>\n

    \n<\/p>

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

    Quadratic equations, Quadratic Functions Many IB Maths Tutors consider quadratic equations as a very important topic of maths. There are the following ways to solve […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11,5],"tags":[],"class_list":["post-311","post","type-post","status-publish","format-standard","hentry","category-cbse-tutors","category-ib-mathematics-tutors"],"_links":{"self":[{"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/posts\/311","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/comments?post=311"}],"version-history":[{"count":0,"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/posts\/311\/revisions"}],"wp:attachment":[{"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/media?parent=311"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/categories?post=311"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/ibelitetutor.com\/blog\/wp-json\/wp\/v2\/tags?post=311"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}