Quadratic Formula

Question

Quadratic Formula

GeeksforGeeks · Accepted Answer

Quadratic Formula is used to find the roots (solutions) of any quadratic equation. Using the Quadratic formula real and imaginary all the types of roots of the quadratic equations are found.The quadratic formula was formulated by a famous Indian mathematician Shreedhara Acharya, hence it is also called Shreedhara Acharya's Formula. It is used to find the solution of the quadratic equation of the form ax2 + bx + c = 0where a &ne; 0, and a, b, and c are real constants.For a Quadratic equation, f(x) = ax2+ bx + c where a is not equal to 0 and a, b, and c are real numbers; the solution of f(x) = 0 is given by Quadratic Formula i.e., Quadratic FormulaThe term under the square root i.e., b2 - 4ac, is called the discriminant, which determines the nature of the roots: discriminant &gt; 0, the equation has two real and distinct roots.discriminant = 0, the equation has exactly one real root (a repeated root).discriminant &lt; 0, the equation has two complex (or imaginary) roots.Table of ContentDerivation of Quadratic FormulaBy Completing the Square TechniqueShortcut Method of DerivationRoots of Quadratic Equation by Quadratic FormulaDiscriminant of a Quadratic EquationNature of Root of Quadratic EquationMaximum and Minimum Value of Quadratic ExpressionWhat is Quadratic Formula used for?Solving Quadratic Equation Using Quadratic FormulaSolved Example on Quadratic FormulaExample: Find the roots of the equation x2 + 5x + 6 = 0.SolutionFor x2 + 5x + 6 = 0, use the quadratic formula:a = 1, b = 5, and c = 6.Substitute into the formula:x = frac{-5 pm sqrt{5^2-4(1)(6)}}{2(1)}x = frac{-5 pm sqrt{25-24}}{2}x = frac{-5 pm 1}{2}The two solutions are: x = &minus;2 and x = &minus;3Derivation of Quadratic FormulaWe can easily derive the quadratic formula, using two methods&nbsp;By Completing the Square TechniqueShortcut Method of DerivationNow let's learn about them in detail.By Completing the Square TechniqueDerivation of the quadratic formula is achieved by using the Completing Square methodLet us take the standard form of a quadratic equation i.e. ax2 + bx + c = 0Dividing the equation by the coefficient of x2, i.e., x2 + (b/a)x + (c/a) = 0Subtracting c/a from both sides: x2 + (b/a)x = -c/aNow, by completing the square method,We have to add a specific constant to both sides of the equation to make the LHS a complete square.Here, we add (b/2a)2 to both sides of the equationx2 + (b/a)x + (b/2a)2 = (-c/a) + (b/2a)2Using a2 + 2ab + b2 = (a + b)2,&rArr; [x + (b/2a)]2 = (-c/a) + (b2/4a2)&rArr; [x + (b/2a)]2 = (b2 &ndash; 4ac)/4a2Taking square root on both sides, [x + (b/2a)] = &radic;[(b2 &ndash; 4ac)] / 2aNow, subtracting b/2a from both sides we getx = [-b &plusmn; &radic;(b2&nbsp;&ndash; 4ac)] / 2aThis is the required quadratic formula.Shortcut Method of DerivationThe quadratic formula is derived by the shortcut method as,The given standard form of a quadratic equation is,ax2 + bx + c = 0Multiply both sides by 4a4a(ax2 + bx + c) = 4a &times; (0)&rArr; 4a2x2 + 4abx + 4ac = 0&rArr; 4a2x2 +4abx = -4acBy completing the square method, add b2 on both sides.4a2x2 + 4abx + b2 = b2 &ndash; 4ac&rArr; (2ax)2 + 2(2ax)(b) + b2 = b2 &ndash; 4ac &nbsp; &nbsp;{We know that, a2 + 2ab + b2 = (a + b)2)}&rArr; (2ax + b)2 = b2 &ndash; 4acTaking square root2ax + b = &plusmn;&radic;(b2 &ndash; 4ac)&rArr; 2ax = -b &plusmn;&radic;(b2 &ndash; 4ac)x = [-b &plusmn;&radic;(b2 &ndash; 4ac)]/2aThis proves the quadratic formula.Roots of Quadratic Equation by Quadratic FormulaA quadratic equation is a two-degree quadratic equation and it has two a maximum of two roots. They can either be real, or imaginary.&nbsp;The roots of the quadratic equation is the value that satisfies the quadratic equation thus, we can say that if the given quadratic equation is, ax2 + bx + c = 0, and if &alpha; is a root of the quadratic equation, then &alpha; satisfies the quadratic equation, i.e., a&alpha;2 + b&alpha; + c = 0. We also define roots of the quadratic equations ax2 + bx + c = &nbsp;as the zeros of the polynomial ax2 + bx + c.We can easily find the roots of the quadratic equation by using the following quadratic formula, bold{x = frac{-b pm sqrt{b^2 &ndash; 4ac}}{2a}}This gives the two roots of the quadratic equation,&nbsp;Taking +ve Signx = [-b + &radic;(b2&nbsp;&ndash; 4ac)] / 2aTaking -ve Signx = [-b - &radic;(b2&nbsp;&ndash; 4ac)] / 2aThus using the quadratic formula we easily get the roots of the quadratic equation.Discriminant of a Quadratic EquationThe discriminant of a quadratic equation ax2 + bx + c = 0 is denoted by D and given by,D = b2 - 4acDiscriminant of a Quadratic Equation is very helpful in determining the nature of the root of quadratic equations.Nature of Root of Quadratic EquationTo find the nature of the roots of the quadratic equation we find the discriminant of the given quadratic equation. The term is called discriminant because it determines the nature of the roots of the quadratic equation based on its sign.There are 3 types in the nature of roots,&nbsp;DiscriminantNature of RootsD &gt; 0 (b2 - 4ac &gt; 0)Real and distinct rootsD = 0 (b2 - 4ac = 0)Real and equal rootsD &lt; 0 (b2 - 4ac = 0)Imaginary Roots.Maximum and Minimum Value of Quadratic ExpressionThe maximum and minimum values for the quadratic equation of the form ax2 + bx + c = 0 can be observed with the help of graphs.If the value of a is positive i.e. (a &gt; 0), the quadratic equation has a minimum value at x = -b/2a i.e., -D/4a.If the value of a is negative i.e. (a &lt; 0), the quadratic equation has a maximum value at x = -b/2a i.e., -D/4a.Where D is the discriminant of the Quadratic Expression.What is Quadratic Formula used for?The quadratic formula is the most used formula in algebra and it is used for solving quadratic equations and finding their roots instantly without performing various steps. It gives the solution of the quadratic equations where the normal solution is not possible. We use this formula for solving various types of problems in algebra, factoring equations, graphing various curves, and others.Solving Quadratic Equation Using Quadratic FormulaFollow the steps given below to solve Using Quadratic FormulaStep 1: Write the given equation in standard form as, ax2+ bx + c = 0Step 2: Carefully note the coefficient from the above equation as, a, b and c.Step 3: Use the Quadratic Formula, x = [-b &plusmn; &radic;(b2&nbsp;&ndash; 4ac)] / 2aStep 4: Put all the values of a, b and c and simplify for x.We can understand this with the help of the following example.Example: Simplify x2 + x - 6Solution:Step 1: Comparing x2 + x - 6 with ax2+ bx + c = 0Step 2: a = 1, b = 1, and c = -6Step 3: Using the quadratic formula,x = [-b &plusmn; &radic;(b2&nbsp;&ndash; 4ac)] / 2aStep 4: Simplifyingx = [-b &plusmn; &radic;(b2&nbsp;&ndash; 4ac)] / 2a&rArr; x = [-1 &plusmn; &radic;(12&nbsp;&ndash; 4(1)(-6))] / 2(1)&rArr; x = [-1 &plusmn; &radic;(4&nbsp;+ 4(1)(6))] / 2(1)&rArr; x = (-1 &plusmn; 5)/2&rArr; x = (-1-5)/2 = -3&rArr; x = (-1+5)/2 = 2Thus, the values of x are x = -3 and x = 2Read More,Quadratic&nbsp;FunctionQuadratic InequalitiesSolving Cubic&nbsp;EquationsSolved Example on Quadratic FormulaExample 1: Write the quadratic function f(x) = (x - 9)(x + 3)in the general form ofax2 + bx + c.Solution:&nbsp;Given, the function as (x - 9)(x + 3)= x2 + 3x - 9x - 27= x2 - 6x - 27This is the required form.Example 2: Find the constants a, b, and c in the general form of equation 4x2 + 5x + 9 = 0.Solution:&nbsp;Given, equation is 4x2 + 5x + 9 = 0....(1)General form of quadratic equation is ax2 + bx + c = 0.Comparing the given equation with general equation we get,a = 4, b = 5, c = 9.Example 3: Write the quadratic function f(x) = (x + 8)(x - 3) in the general form of ax2 + bx + c.Solution:Given, the function as (x + 8)(x - 3)&nbsp;= x2 - 3x + 8x - 24= x2 + 5x - 24 [Which is the general form]Example 4: Find the roots of equation 2x2 - 4x + 2 = 0.Solution:&nbsp;Here a = 2, b = -4, c = 2, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.&rArr; Discriminant = b2 - 4ac = (-4)2 - 4(2)(2)&nbsp;&rArr; Discriminant = 16 - 16 = 0 [which is equal to zero]&nbsp;So, it has real and equal roots.Roots = (&minus;b &plusmn; &radic;(b2 &minus; 4ac)) / 2a&rArr; Roots = {-(-4) &plusmn; &radic;(0) } / 2(2)&rArr; Roots = 4 / 4 = 1Thus, 1 is the root of the equation.Example 5: Find the roots of equation 4x2 - 3x + 3.Solution:&nbsp;Here a = 4, b = -3, c = 3, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.Discriminant = b2 - 4ac&nbsp;&rArr; Discriminant = (-3)2 - 4(4)(3)&nbsp;&rArr; Discriminant = 9 - 48 = -39 &nbsp;[which is negative]So, it has complex roots.Roots = {&minus;b &plusmn; &radic;(b2 &minus; 4ac)] / 2a&rArr; Roots = (-(-3) &plusmn; &radic;-39 / 2(4) )&rArr; Roots = (3 &plusmn; 39i)/8&nbsp;Thus, (3 + 39i)/8 and (3 - 39i)/8 are the roots of the quadratic equation.Example 6: Find the roots of the quadratic equation 6x2 - 8x + 2 = 0.Solution:Here a = 6, b = -8, c = 2, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.Discriminant = b2 - 4ac&nbsp;&rArr; Discriminant = (-8)(-8) - 4(6)(2)&nbsp;&rArr; Discriminant = 64 - 48 = 16 [which is positive]&nbsp;So, it has real and distinct roots.x = (-b &plusmn; &radic; (b&sup2; - 4ac) )/2a&rArr; x = (-(-8) &plusmn; &radic;(16) / 2(6)&rArr; x = (8 &plusmn; &radic;16) / 12&rArr; x = &nbsp;(8 &plusmn; 4)12Taking +ve sign, we get x = 1, and&nbsp;Taking -ve sign, we get x = 1/3.Thus, 1/3 and 1 are the roots of the equation.Quadratic Formula -