Solution of all math equations and square method

This presentation help student in the all math equations to solve them easily.

1. PRESENTED TO: SIR ALI RAZA SUBJECT:EXPLORING QUANTITATIVE SKILLS(MATHS1126)​ PRESENTED BY:GROUP 2 QUADRATIC EQUATION IN ONE VARIABLE & PRACTICAL SNERIO INVOLVING EXPRESSIONS GROUP PARTICIPANTS: RABIA SANI, MEHAK FATIMA KISHAAF MAQBOOL, AREESHA RAZA

2. MAIN POINTS TO DISCUSS What is quadratic equation in one variable?​ How it differs from linear equation?​ Forms of quadratic equation ( standard & pure).​ Solution of quadratic equation​ Types of solution​ Practical snerio involving expressions

3. QUADRATIC EQUATION DEFINITION : An equation which contains the square of variable quantity, but no high power , is called a quadratic equation OR an equation of the second degree. DIFFERENCE BETWEEN LINEAR AND QUADRATIC EQUATION: It differes from linear in such a way that : T he linear euation is with highest power 1 of its variable wherease quadratic equation is with variable of second degree. T here is single answer( 1 root) in solution set of linear equation ,however, the solution set of quadratic equation contains two possible values(2 roots). FORMS OF QUADRATIC EQUATION: Standard form: i.e. ax 2 + bx + c = 0 , (where a, b, and c are constants)(a is not 0) Pure form: i.e. ax 2 +c=0, (where b is 0)

4. SOLUTION OF QUADRATIC EQUATION Rewrite in standard form i.e. ax 2 + bx + c = 0 (splitting) Find factors of (c)(a) that add to “b” (expanding) Grouping & solution (grouping) FACTORIZATION Rewriting in formate i.e. ax 2 + bx= -c Complete the square: square=(b/2)2 add on bothsides Write in whole square and solve by taking sq roots COMPLETING SQUARE METHOD Rewrite in standard form Put values in quadratic formula: x= (-b±√(b²-4ac))/(2a) Solve USING QUADRATIC FORMULA 01 02 03

5. FACTORIZATION METHOD RULES TO SOLVE IT: Rewrite in standard form i.e. ax 2 + bx + c = 0(splitting) Find factors of (c)(a) that add to “b” (expanding) Grouping & solution(grouping) EXAMPLE NO. 1 Solve the equation, x2−3x−10=0 Factor the left side: (x−5)(x+2)=0 Set each factor to zero: x−5=0  or  x+2=0 Solve each equation: x=5  or  x=−2 The solution set is ss= {5,−2} .

6. COMPLETING SQUARE METHOD RULES TO SOLVE: Rewriting in formate i.e. ax 2 + bx= -c Complete the square: square=(b/2)2 add on bothsides Write in whole squareand solve by taking sq roots EXAMPLE no.2 x 2 + 4x – 5 = 0 b = 4, c = -5 (x + b/2)2 = -(c – b2/4) So, [x + (4/2)]2 = -[-5 – (42/4)] (x + 2)2 = 5 + 4 ⇒ (x + 2)2 = 9 ⇒ (x + 2) = ±√9 ⇒ (x + 2) = ± 3 ⇒ x + 2 = 3, x + 2 = -3 ⇒ x = 1 , -5 ss={1, -5}

7. DERIVATION OF QUADRATIC FORMULA The quadratic formula, which is used to find the solutions of a quadratic equation in the form ax^2 + bx + c = 0, can be derived by completing the square. Here's a step-by-step derivation: Starting with the equation: ax^2 + bx + c = 0 Divide both sides of the equation by 'a' x^2 + (b/a)x + c/a = 0 Move the constant term 'c/a' to the right side x^2 + (b/a)x = -c/a To complete the square on the left side of the equation, you need to add and subtract (b/2a)^2. This is because (b/2a)^2 is the term that, will make the left side a perfect square trinomial. x^2 + (b/a)x + (b/2a)^2 = -c/a +(b/2a)^2 left side can be factored as a perfect square: (x + b/2a)^2 = -c/a + (b/2a)^2 Take the square root of both sides: x + b/2a = ±√(-c/a + (b/2a)^2) Solve for 'x' by isolating it on the left side: x = -b/2a ± √(-c/a + (b/2a)^2) Simplify the right side: x = (-b ± √(b^2 - 4ac)) / (2a)

8. USING QUADRATIC FORMULA RULES TO SOLVE IT: Rewrite in standard form Put values in quadratic formula: x= (-b±√(b²-4ac))/(2a) Solve Example no 3; 2x 2 - 5x + 3 = 0 where a=2,b=-5,c=3 putting values in formula, x= (-b±√(b²-4ac))/(2a) x = (-(-5) ± √((-5)^2 - 4(2)(3))) / (2(2)) x = (5 ± √(25 - 24)) / 4 x = (5 ± √1) / 4 Calculate the two possible solutions: x₁ = (5 + 1) / 4 = 6 / 4 = 3/2 x₂ = (5 - 1) / 4 = 4 / 4 = 1 ss={1, 3/2}

9. PRACTICAL SNERIO INVOLVING EXPRESSIONS

10. EXAMPLE NO. 1 You are asked to built a rectangular field with area=7000m 2 .The length of the fieldshould be 30 mmore than its width.How will you build the field? Solution: Let suppose that,width= x then according to the given condition; length =x+30 for area of a rectangle, A=(l)(b) 7000=(x+30)(x) 7000=x 2 +30x l=x+30m x 2 +30x-7000=0 A=7000m 2

11. Three less than a certain number multiplied by 9 less than twice the number is 104. Find the number. SOLUTION Let the required number= x. Then 3 less then that =x-3 and nine less than twice the number=2x-9 According to the given condition: (x-3)(2x-9)=104 2x 2 -15x+27=104 2x 2 -15x-77=0 EXAMPLE NO.2

12. The length of a base and the corresponding height of a triangle are (x+3) and (2x-5) respectively.given the area of the triangle is 20cm 2 .Find the value of x. We are given that; Area=20cm 2 Base=x+3 Height=2x-5 as Area of a triangle =1/2(b)(h) putting values we get: 20=1/2(x+3)(2x-5)=(2x 2 -5x+6x-15)/2 20=2x 2 +x-15/2 20 X 2=2x 2 +x-15 0=2x 2 +x-40-15 2x 2 +x-25=0 b=x+3 EXAMPLE NO.3 h=(2x-5)