Page 1 : Downloaded from https:// www.studiestoday.com, , Gungrene DAB ODR, , , , , , , , , , , , , , , , , , 5.1 | Introduction :, , , , , , , , An equation with one variable, in which the highest power of the variable is two,, is known as quadratic equation,, , For example :, (i) 3x27 + 4x4+7=0 (iii) 2x? - 50 = 0, (ii) 4x? + 5x = 0 (iv) x? = 4, etc., , , , . The standard form of a quadratic equation is ax* + bx + c = 0, where a, b and c are all real, numbers and a # 0., , e.g. equation 4x? + 5x — 6 = 0 is a quadratic equation in standard form., , 2, Every quadratic equation gives two values of the unknown variable used in it and ee, values are called roots of the equation., , 3. Discriminant : For the quadratic equation ax* + bx + c = 0, a # 0; the expression., - 4ac is called discriminant and is, in general, denoted by the letter ‘D’., , Thus, discriminant D = b? - dac., , 4. If a quadratic equation contains only two terms one square term and one first power term, of the unknown, it is called adfected quadratic equation., , For example : (i) 4x2 + 5x=0 (ii) 7x - 3x = 0, etc., , 5. If the quadratic equation contains only the square of the unknown, it is called pure, quadratic equation., , For example : (i) x7 =4 (ii) 3x7 - 8 = 0, ete., , , , , , , , 5.2 | To examine the nature of the roots :, , , , , , , , Examining the roots of a quadratic equation means to know the type of its roots, ie. whether they are real or imaginary, rational or irrational, equal or unequal., , The nature of the roots of a quadratic equation depends entirely on the value of, its discriminant b? — 4ac., , If for a quadratic equation ax? + bx + c = 0; where a, b and c are real numbers, and a # 0, then discriminant :, , (i) b* - 4ac = 0 = the roots are real and equal., (ii) 6? - dac > 0 = the roots are real and unequal., (iii) b? - 4dac < © = the roots are imaginary (not real)., , , , , , 49°, Downloaded from https:// www.studiestoday.com

Page 2 : Downloaded from https:// www.studiestoday.com, , 1. Every number, whether it is rational or irrational, is a real number. i.e., (i) every rational number is a real number and, (ii) every irrational number is also a real number., , 2. Square root of a negative number is an imaginary number., , , , , , ut solving, examine the nature of the roots of the equations :, -6x+7=0 (i) x2 +6x+9=0 — (ili) 2x2 + 6x +3 =, , , , , , Solution :, , (i) Comparing given quadratic equation 5x? — 6x + 7 = 0 with equation ax? + bx + c = 0;, we get:a=5,b=-6 andc=7., , => Discriminant = b? - 4ac = (-6)*-4 x 5x7, = 36 — 140 = -104; which is negative., Since, a, b and c are real numbers; a # 0 and b? — 4ac < 0., .. The roots are not real i.e. the roots are imaginary. Ans,, , (ii) Comparing quadratic equation x? + 6x + 9 = 0 with ax* + bx + c = 0; we get :, a=1,b=6andc=9, , > b - 4ac= (6)? -4x1x9=36-36=0, Since; a, b and c are real numbers; a # 0 and b? — 4ac = 0., .. The roots are real and equal. Ans., , (iii) Comparing 2x? + 6x + 3 = 0 and ax? + bx + c, we get: a= 2, b= 6 andc =3, B? - 4ac= (6)? -4x2x3, = 36 — 24 = 12; which is positive., Since; a, b and c are real numbers; a # 0 and b? — 4ac > 0., .. The roots are real and unequal. Ans., , , , , , , , , lue of ‘m’, if the roots of the following quadratic equation are eq, Fm)? + (m+ 1x +1 = 0. aged, , Solution :, For the given equation (4 + m)x? + (m + 1)x + 1=0;, a=4+m,b=m+landc=1, , Since, the roots are equal, , BP -4ac=0=> (m+1P-44+m)x1=0, => m+2m+1-16-4m=0, => m —2m-15 = 0, On solving, we get: m=5 or m=-3 Ans., , , , Downloaded from https:// www.studiestoday.com

Page 3 : Downloaded from https:// www.studiestoday.com, , Gases) ——_—_, , 1. Without solving, comment upon the nature of | 3. The equation 3x2 - 12x + (n - 5) = 0 has, roots of each of the following equations : equal roots. Find the value of n., @ 72-9x+2=0 ii) Ge - 13x+4=0, , 4. Find the value of ‘m’, if the following equation, (iii) 25x? - 10x + 1 = 0 (iv) 2 + 2¥3x-9=0 sea, , has equal roots :, , (vy) @-ax-B=0 (vi) 2x27 + 8x+9=0 (m — 2)x2 - (5 + m)x + 16 =0, 2. Find the value of ‘p’, if the following quadratic | 5 Find the value of k for which the equation, equations have equal roots : 3x2 — 6x + k = 0 has distinct and real root., @) 42-(p-2x+1=0 [2015], , (ii) 22 +p - 3)x +p =0 [2013], , , , , , , , Solving quadratic equations by factorisation :, , Clear all fractions and brackets, if necessary., Transpose all the terms to the left hand side to get an «, the form ax? + bx +c = 0., , Factorise the expression on the left hand side., Put each factor equal to zero and solve., , , , , , , , , , , ict Rule : Whenever the product of two expressions i, expressions is zero., ‘Thus, if (x + 3) (¢-2) =, , x+3=0, o x-2=0, = 2), , , , , , >, >, , , , , , 2x? — 7x = 39, 2x -7x-39 =0 [Expressing as ax? + bx +c = 0), 2x2 - 13x + 6x - 39 = 0 [Factorising the left hand side], , g, puusk, =, , x (2x — 13) +3 (2x - 13) = 0, (2x - 13) (x + 3) = 0, , 2x-13=0,orx+3=0 [Zero Product Rule], are Borxe3 Ans., Gi) x2=Sx => x -5x= 0, = x(x-5)=0, > x=0,orx-5=0, => x=0,orx=5 Ans., (ii) x2=16 = 2-16=0 | Alternative method :, => (+4) (@-4)=0 x= 16, => x+4=0,orx-4=0 | = xat+4, |, , x=-4,orx= 4 Ans. => x=4orx=—-4 Ans., , JU, , Downloaded from nttba www.studiestoday.com

Page 4 : Downloaded from https:// www.studiestoday.com, , , , , , , , , , e x x-1 2, Solve : —— + ——=2-., " x-1 x 29, Solution :, x x-1, — 12> 2. 2x-1 x, x24+(x-)? _ 5, om x(x-1) "9, x? +22 -2x4+1) = 5(2-x), => 4x -4x+2 = 5x2 - 5x, => w+x+2 = 0, > w-x-2=0 [Changing the sign of each term], = -2@+)=0 [On factorising], => x-2=0,orx+1=0 [Zero Product Rule], => x=2, or x = -1 Ans., , , , , , the quadratic equation whose solution set is {-2, 3}., , , , , , , , Solution :, Since, solution set = {-2, 3}, = Roots are -2 and 3, > x=-2,0r x=3, x+2=0,orx-3=0, (x + 2) @- 3) =0, x -3x+2x-6=0, , Quugy, , x*—x—6 =0; which is the required quadratic equation. Ans., , , , , , substitution x = 3y + 1 to solve for y, if 5(3y + 1/2 + 6(3y + 1) - 8=0, |, , , , , , , , , , Solution :, 5(3y +1)? + 63y + 1)-8 = 0, , => 5x27 +6x-8 = 0 [Putting 3y + 1 = x], , > (x + 2) (5x-4) = 0 [On factorising], 4, , => x=-2,orx = >, , 5, Whenx=-2 => 3y+1l=-2>y=-l, 1, , 4 4, and, whenx=— => 3y+1l=—-— >y=-—, 5 Ps 5 * 15, , 1, “ysl =-—, y 9» Ory 15, , Downloaded from riteed A wcotianetoa icon, , Ans,

Page 5 : Downloaded from https:// www.studiestoday.com, , , , , , Without solving the quadratic equation 3x2 - 2x - 1 = 0, find whether x., a solution (root) of this equation or not., , , , , , , , Solution :, Substituting x = 1 in the given equation 3x? — 2x - 1 = 0,, we get : 301)? -2x1-1=0, > 3-2-1 = 0; which is true., x = 1 is a solution of the given equation 3x? - 2x - 1 = 0 ; Ans,, , Oo Without solving equation x? - x + 1 = 0; find whether x = -1 Is a root of |, equation or not., , , , , , Solution :, Substituting x = —1 in the given equation x? - x + 1 = 0,, we get : (-1)?-(-1)+1=0, ie. 1+1+1=0 => 3 = 0; which is not true., x =-1 is not a root of the given equation x? —- x + 1=0, , , , Find the value of k for which x = 2 is a root (solution) of equation, kx? + 2x - 3 =0., , , , , , Solution :, Substituting x = 2 in the given equation kx? + 2x — 3 = 0; we get :, K(2)?+2x2-3=0, y, , , , , , , , , , = 4k+4-3=0 = SS Ans., Ly is a, te If x = 2 and x = 3 are roots of the equation 3x? - 2mx + 2n = 0; fin, |._-values of m and n. et, , , , , , Solution :, x = 2 is a root of the equation 3x? — 2mx + 2n = 0, , => 3(2)? -2mx2+2n=0, > 12-4m+2n=0, > -4m+2n=-12 ite 2m-n=6 renee I, x = 3 is a root of the equation 3x? — 2mx + 2n = 0, > 3(3)? — 2m x 34+ 2n=0, => 27-6m+2n=0, > — 6m + 2n = -27 ie. 6m-2n=27 sien, On solving equations I and II, we get :, m=T75 and n=9 Ans., , Downloaded from https:/7 www.studiestoday.com