# Mathematics - Class 10 / Grade 10

- Solution of a Quadratic Equation by Factorization
### Solution of a Quadratic Equation by Factorization

**Tags:**Solution of a Quadratic Equation by Factorisation practice page for class 10, Solving quadratic equation by factorization method, Zeroes of the Quadratic Polynomial, Roots of the Quadratic Equation, Solving quadratics by factoring notes for Grade X, Solve quadratic equations by factoring worksheet PDF for 10^{th}standard, How Do You Solve a Quadratic Equation by Factoring? Factoring Quadratics examples for tenth gradeSolution of a Quadratic Equation by Factorization

A real number ‘α’ is

__called a root of the quadratic equation__**a**x^{2}+**b**x +**c**= 0, a ≠ 0 if a**α**^{2}+**b**α +**c**= 0. In other word x = α is a solution of the quadratic equation, or that x = α satisfies the quadratic equation.__Zeros of the quadratic polynomial__**a**x^{2}+**b**x +**c**and__the roots of the quadratic equation__**a**x^{2}+**b**x +**c**= 0 are the__same__. We have observed that a quadratic polynomial can have at most two zeros.We can find the roots of quadratic equation by factorizing quadratic polynomials by splitting their middle terms into two linear factors and equating each factor to zero.

For quadratic equation be

**a**x^{2}+**b**x +c = 0; a ≠ 0.The quadratic polynomial**a**x^{2}+**b**x + c be expressed as the product of two linear factors, say (px + q) and (rx + s), where p, q, r, s are real numbers such that p ≠ 0 and r ≠ 0. Then,ax

^{2}+ bx + c = 0(px + q) (rx + s) = 0

px + q = 0 or, rx + s = 0

Solving these linear equations, we get the possible roots of the given quadratic equation as x = -p/q and x = -s/r

**Example 1:**Find the roots of the equation x^{2}+ 6x + 5 = 0**Solution:**Here, a x c is 1×5 = 5 and b is 6So we have to look for two numbers that multiply together to make 5, and add up to 6.

Factors of 5 include 1 and 5. (1 and 5 add to 6, and 1×5=5)

So, x

^{2}+ 5x + 1x + 5 = 0 (split the middle term 6x as 5x + 1x [as (5x) X (1x) = 5x^{2}]x(x + 5) + 1(x + 5) = 0 (take out common from first two term and common from last two term)

(x + 5) (x + 1) = 0 (write the common factor and other factor)

x + 5 = 0 or, x + 1 = 0 (equate each factor to zero)

x = -5 or x = -1

Thus, x = -5 and x = -1 are two roots of the equation x

^{2}+ 6x + 5 = 0**Example 2:**Solve the quadratic equation by factorization method: x^{2}+ 2√2x – 6 = 0**Solution:**x^{2}+ 2√2x – 6 = 0x

^{2}+ 3√2x – √2x - 6 = 0 [as; 3√2x - √2x = 2√2x and (3√2x) X (√2x) = (3 x 2)x^{2}= 6x^{2 }]x(x + 3√2) - √2(x + 3√2) = 0

(x + 3√2) (x - √2) = 0

x + 3√2 = 0 or x - √2 = 0

x = - 3√2 or x = √2

Thus, x = - 3√2 and x = √2 are two roots of the given equation.

**Example 3:**Solve the quadratic equation by factorization method: 4/x -3 = 5/(2x+3)**Solution:**Take the LCM of the denominator of the fractions.(4-3x)/x = 5/(2x+3)

(4-3x) (2x+3) = 5x

12 – x – 6x

^{2}= 5x6x

^{2}+ 6x -12 = 0x

^{2}+ x – 2 = 0x

^{2}+ 2x –x - 2 = 0x(x+2) – (x+2) = 0

(x+2) (x-1) = 0

x+2 = 0 or x-1 = 0

x = -2 or x = 1

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