NCERT solutions for class 10

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2. https://www.entrancei.com/ QUADRATIC EQUATIONS 1. Quadratic Polynomial A polynomial in the form p (x ) = and x is a real variable is called a quadratic polynomial. 2 , where a 0 and a, b, c are real numbers + + ax bx c 2. Quadratic Equation An equation p (x ) = 0 where p (x ) is a quadratic polynomial is called a quadratic equation, i.e., 0 = + + c bx ax , where a 0. 2 2 2 − x + + x − 2 x 3 , 7 e.g. 8 x 19 3. 4. Pure Quadratic Equation Equation of type 2 +c = is known as pure quadratic equation. ax 0 Roots of a Quadratic Equation The values of variable satisfying the given quadratic equation are called its roots. For Example: Determine whether the given values of x are the roots of the given equation or not: 2 + + = = , 1 − = − x 6 x 5 ; 0 x x 5 2 + x + Consider = …(i) p (x ) x 6 5 Substituting x = –1 in (i), we get 2 ) 1 − +  ) 1 − + − ) 1 = ( 6 ( 5 ( p = 1 – 6 + 5 = − = . 6 6 0 Hence x = −1 is the root of given equation. Again substituting x = –5 in (i), we get 2 − +  − + (− = ( 5 ) 6 ( 5 ) 5 p 5 ) = 25 – 30 + 5 = 30 – 30 = 0 Hence x = −5 is the root of given equation.  Methods to Find Roots of a Quadratic Equation (i) Discriminant Method 5. x = –1 and x = –5 are the roots of given equation. 2 + + = Let quadratic equation be ax bx c 0 2− = D b ac 4 Step 1: Find ; D is called the discriminant of the equation. − + − − b D b D  Step 2: (a) If roots are given by x = . D , 0 , 2 a 2 a − b = (b) If equation has equal roots and root is given by x = . D , 0 2 a https://www.entrancei.com/ncert-solutions-class-10-maths

3. https://www.entrancei.com/  (c) If equation has no real roots. D , 0 For Example: Using quadratic formula solves the quadratic equation: 2 2 2 2 2 + − − = p x ( p q ) x q 0 We have, 2 2 2 2 2 + − − = p x ( p q ) x q 0 2 + + = Comparing the equation with ax bx c , 0 we have 2 2, 2 2 = = − = − and a p b p q c q 2 2 2 2 2 2 2 2 2 2 2  = − = − −   − = − + D b 4 ac ( p q ) 4 p q ( p q ) 4 p q 4 4 2 2 2 2 2 2 2 = + − + = +  p q 2 p q 4 p q ( p q ) 0 [ square of any real number is non-negative] So, the given equation has real roots given by 2 2 2 2 2 2 − + − − + + b D ( p q ) p ( p q ) 2 q q  = = = = 2 2 2 2 a 2 2 p p 2 2 2 2 − − − − − + b D ( p q ) p ( p q )  = = = − and, 1 2 2 a 2 2 q Hence the roots are and -1 2 p (ii) Factorisation method 2 + + Step 1: Resolve the quadratic polynomial factors. i.e., numbers and A, C 0 Step 2: Put the linear factors equal to zero and get the values of x. (Ax + B)(Cx + D) = 0 into the product of linear where A, B, C and D are real ax )( Cx bx ) D c 2 + + = + + ax bx c ( Ax B Ax + B = 0, Cx + D = 0 − − B D = = x , x are the two roots of quadratic equation. A C For Example: 2 2 2 2 2 + − − = Solve by factorisation: a b x b x a x 1 0 2 2 2 2 2 + − − = a b x b x a x 1 0 2 2 2  ) 1 + ( 1 − ) 1 + = b x ( a x a x 0 ( ) 2 2  + − = ( a x ) 1 b x 1 0 2 2 ) 1 + = ) 1 − = Either or ( a x 0 ( b x 0 2 2  = − = or a x 1 b x 1 https://www.entrancei.com/ncert-solutions-class-10-maths

4. https://www.entrancei.com/ 1 1  = − x = or x 2 2 b a 1 1 = − Hence, are the required solutions x , 2 2 a b (iii) Completion of square method Let quadratic equation be 2 + + = ax bx c 0   b c 2 2 + + = Step 1: Make the coefficient of x equal to 1, i.e., a x x 0     a a b c 2  + + = x x 0 a a Step 2: Add and subtract the square of half of coefficient of x,  − + + a a Step 3: Express in the form of a perfect square added with a constant,     2 2 b b b c   2 + = i.e., x x 0     2 2 a 4 4 a  2 2   b b c  + − + = i.e., x 0     2 2 a a    4 a 2 2 2 −   b b c b 4 ac + = − = Step 4: x     2 2 2 a a 4 a 4 a 2 − b b 4 ac + =  Step 5: x 2 2 a 4 a 2 − − b b 4 ac =  Step 6: x 2 2 a 4 a For Example: Solve the equation We have, x  2 − ) 1 + + = by the method of completing the square. x ( 3 x 3 0 3 2 − ) 1 + − + + = = ( x 3 0 − 2 x ( 3 ) 1 x 3 2   + 3 1     Adding on both sides we have 2 2 2       + + + 3 1 3 1 3 1             2  − + = − + x 2 x 3 2 2 2 2   + + + 3 1 3 1 2 3      − = − + x 3 2 4 − + + + 4 3 3 1 2 3 = 4 https://www.entrancei.com/ncert-solutions-class-10-maths

5. https://www.entrancei.com/ + − 3 1 2 3 = 4 2 2     + − 3 1 3 1          − = x 2 2 + − 3 1 3 1  − =  x 2 2 − + 3 1 3 1 2 3  = + = = x 3 2 2 2 + − 3 1 3 1 − = − or x 2 2   − + 3 1 3 1 2     = − + = = x 1 2 2 2  = x , 3 1 Hence, the roots are 3 and 1. 6. Nature of Roots Let quadratic equation be (i) If , 0  D then equation has real and unequal roots. 2 2−  + + = = , and , a , 0 ax bx c 0 D b 4 ac For Example: Find the values of k for which the given equation has real and unequal roots: 2 − x − = kx We have 6 2 0 2 − x a − , = = Here, 6 kx 6 = 2 b 0 − and c = –2 k 2 2  = − = − −   − = + D b 4 ac ( 6 ) 4 k 2 36 8 k The given equation will have real and unequal roots, if − 36 9   +    36 −     − D 0 36 8 k 0 8 k k k 8 2 = (ii) If then equation has equal and real roots. D , 0 For Example: Find the values of k for which the following equation has equal roots: 0 2 ) 12 ( 2 ) 12 ( = + − + − x k x k 2 We have, 2 − + − + = ( k 12 ) x 2 ( k 12 ) x 2 0 = − = − Here, and c = 2 a k 12 , b 2 ( k 12 ) 2 2  = − = − − −  D b 4 ac 4 ( k 12 ) 4 ( k 12 ) 2 https://www.entrancei.com/ncert-solutions-class-10-maths

6. https://www.entrancei.com/  = − − } 2 − D 4 ( k 12 ) {( k 12 )  = − − D 4 ( k 12 ( ) k 14 ) The given equation will have equal roots, if  k– 12 = 0 or k– 14 = 0 − − = = 4 ( k 12 ( ) k 14 ) 0 D 0 k = 12 or, k = 14  (iii) If (iv) If a, b, c are rational, rational. (v) If a, b, c are rational, D > 0 and not a perfect square, then roots are real, unequal and conjugate irrational, i.e., if one root is then equation has no real roots.  D D , 0 and perfect square, then roots are real, unequal and 0 p + p − then other is . q q 7. 8. Condition for real roots 0  D i.e., b 2 − ac  4 . 0 Results to remember     b b −    a   (i) if and if ax b 0 x x a 0 a 0  x  or a − x  b  b    a 2 2 x  x = a x   − − − − −  =     b ) b )  − (ii) (iii) (iv) (v) (vi) a x x x ( ( a a a a a 0 0 0 a 2 2 = a a a or  a x 2 2 − − −    x  b or x x x )( )( x x , 0 , 0 a b   a For Example: 2 + kx + = Find the values of k for which the equation has no real roots. x 5 16 0 2 + kx + = The given equation is x 5 16 0 2 + + = ax bx c , 0 Comparing the given equation with we have a = 1, b = 5k, c = 16 2 2 2 = − = −   = − D b 4 ac ( 5 k ) 4 1 16 25 k 64 The given equation will have no real roots if D < 0 2  −  25 k 64 0   64 2  −  25   k   0 25 64 2 −   k 0 [If ab < 0 and a > 0, then b < 0] 25 8 8 2 2 −a  −    x , 0 k [If then –a < x < a] 5 5 9. Applications of Quadratic Equations Step 1: Change the word problems into the form of quadratic equations according to the given conditions. Step 2: Solve the quadratic equations. https://www.entrancei.com/ncert-solutions-class-10-maths

7. https://www.entrancei.com/ For example: Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic? 480 Let the x students planned for picnic, then share of each student = Rs. x 480 − x Share of each student when 8 students failed to go = Rs. 8 According to question 480 − x 480 480 x 480 − x 480 x − = 10 8 x − − ( x 8 )  = 10 x ( 8 ) − +  480 x x 480 8 = 10 − ( x 8 ) 3840 2 − x x 10 x 10  = 1 8 − − 2   = − = 80 x 3840 2 10 x 80 x 3840 0 2  − x − = x 8 384 0 2    24 students planned for picnic Students went for the picnic = 24 – 8 = 16. + − − = x x(x + 16) – 24(x + 16) = 0 (x + 16)(x– 24) x = 24, – 16 16 x 24 x 384 0 (Not possible) https://www.entrancei.com/ncert-solutions-class-10-maths

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