### Solution of a Quadratic Equation by Completion of Squares / Solving Quadratics using the quadratic formula

**Tags:**Solving Quadratic Equations Completing the Square Examples for 10^{th}grade, Solving Quadratics by Completing the Square Worksheet PDF for tenth standard, What is completing the square method?, How to Solve by Completing the Square, Completing the square general formula Exercises for class X, Completing the square problems Practice Page for Grade 10, Solving quadratics using the quadratic formula, Who first derived the quadratic formula?, Nature of the Roots of a Quadratic Equation, What is discriminant?, What is real and distinct roots? What are equal roots?Solution of a Quadratic Equation by Completion of Squares

If we have to solve any quadratic equation, we would probably switch to factorization method. But in some cases, it is not convenient to solve quadratic equations by factorization method.

For example, consider the equation

**x**. In order to solve this equation by factorization method we will have to split the coefficient of the middle term 4 into two integers whose sum is 4 and product is 2. Clearly, this is not possible in integers. Therefore,^{2}+ 4x + 2 = 0__we cannot solve this equation by using factorization method__.We can convert any quadratic equation to the form

**(x + a)**and can easily find its roots. Let us consider the equation^{2}– b^{2}= 0**3x**. Here coefficient of x^{2}– 5x + 2 = 0^{2}is not a perfect square. So, we multiply the equation throughout by 3 to get 9x^{2}– 15x + 6 = 0Now we have, 9x

^{2}– 15x + 6= (3x)

^{2 }– 2 x (3x) x 5/2 + 6= (3x)

^{2 }– 2 x (3x) x 5/2 + (5/2)^{2}– (5/2)^{2 }+ 6= (3x-5/2)

^{2 }– 25/4 + 6= (3x-5/2)

^{2 }– ¼(3x-5/2)

^{2 }= ¼3x-5/2 = 1/2 or 3x-5/2 = -1/2

3x=1/2+5/2 or 3x=-1/2+5/2

3x=6/2 or 3x = -4/2

x=3/3 or x=-2/3

x=1 or x = -2/3

Here is a formula for finding the roots of any quadratic. It is proved by completing the square. In other words, the quadratic formula completes the square for us.

__This method is popularly known as Sridharacharya’s formula__as it was first given by an ancient Indian mathematics Sridharacharya around 1025 A.D**Quadratic Formula**

Thus, if b

^{2}– 4ac ≥ 0, then the quadratic equation**a**x^{2}+**b**x +**c**= 0 has two roots α and β given by**Nature of the Roots of a Quadratic Equation**It is clear that the roots of the quadratic equation

**a**x^{2}+**b**x +**c**= 0, a ≠ 0__are real__if b^{2}– 4ac ≥ 0 and if b^{2}– 4ac < 0, the equation will have__no real roots__. In case b^{2}– 4ac = 0 then the two roots are__equal or coincident__each being equal to –b/2a.It is clear that the nature of the roots of the quadratic equation**a**x^{2}+**b**x +**c**= 0, a ≠ 0 depends upon the value of the expression**b**. This expression is generally called the^{2}– 4ac**discriminant**and is denoted by**D**.**Points To Be Remembered**Let

**a**x^{2}+**b**x +**c**= 0, a ≠ 0 be a quadratic equation, then D = b^{2}– 4ac is called the discriminant of the equation.**Example 1:**Determine whether the given quadratic equations have real roots and if so, find the roots. 9x^{2}+ 7x – 2 = 0**Solution:**The given equation is 9x^{2}+ 7x – 2 = 0.Here, a = 9, b = 7 and c = -2.

Therefore D = b

^{2}– 4ac = 7^{2}– 4x9x-2 = 49 + 72 = 121 > 0So, the given equation has real roots, given by

**Example 2:**Find the value of k for which the given equation has real and equal roots.2x

^{2}– 10x + k = 0**Solution:**The given equation is 2x^{2}– 10x + k = 0Here, a =2, b=-10 and c=k

Therefore, D = b

^{2}– 4ac = (-10)^{2}– 4 x 2 x k = 100 – 8kThe given equation will have real and equal roots if D = 0

100 – 8k = 0

k = 100/8 = 25/2

**Example 3:**If -4 is a root of the quadratic equation x^{2}+ px – 4 = 0 and the quadratic equation x^{2}+ px + k = 0 has equal roots, find the value of k.**Solution:**Since -4 is a root of the equation x^{2}+ px – 4 = 0Therefore (-4)

^{2}+ p x -4 – 4 = 0 (As root always satisfies the equation)16 – 4p – 4 = 0

4p = 12

P = 12/4 = 3

The equation x

^{2}+ px + k = 0 has equal roots.Therefore Discriminant = 0

P

^{2}– 4k = 09 – 4k = 0

K = 9/4

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