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Quadratic Equations

    Solution of a Quadratic Equation by Factorisation

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10th class > Quadratic Equations > Solution of a Quadratic Equation by Factorisation

Solution of a Quadratic Equation by Factorisation

Consider the quadratic equation 2x2 – 3x + 1 = 0. If we replace x by 1 on the LHS of this equation, we get 2×12 – (3 × 1) + 1 = 0 = RHS of the equation.

We say that 1 is a root of the quadratic equation 2x2– 3x + 1 = 0. This also means that is a zero of the quadratic polynomial 2x2 – 3x + 1 .

In general, a real number α is called a root of the quadratic equation ax2 – bx + c = 0, a ≠ 0 if a α2+bα + c =0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation.

Note that the zeroes of the quadratic polynomial ax2 – bx + c and the roots of the quadratic equation ax2 – bx + c = 0 are the same.

You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes. So, any quadratic equation can have atmost two roots.

Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation.

Let us first split the middle term – 5x as –2x –x [because (–2x) × (–3x) = 6x2 = 2x2 × 3 ].
So, 2x2 – 5x + 3 = 2x2– 2x – 3x + 3 = 2x (x – 1) (x – 1) = (2x – 3)(x – 1)
Now, 2x2 – 5x + 3 = 0 can be rewritten as (2x – )(x – 1) = .
So, the values of x for which 2x2 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0
i.e., either 2x – 3 = or x – 1 = .
Now, 2x – 3 = 0 gives x = and x – 1 = 0 gives x =
So, x=32 and x = 1 are the solutions of the equation.
In other words, 1 and 32 are the roots of the equation 2x2 – 5x + 3 = 0.

Verify that these are the roots of the given equation.

Note that we have found the roots of 2x2 – 5x + 3 = 0 by factorising 2x2 – 5x + 3 into two linear factors and equating each factor to zero.

Find the roots of the quadratic equation 6x2−x−2=0.

6x2−x−2=0

  • We have 6x2 - x – 2 and we split the -x term as 6x2 + x - x – 2
  • Divide the eq into two parts and split the terms 6x2+ 3x as x (2x + 1) and split the terms -4x -2 as – (2x + 1)
  • Therfore,(3x – )(2x + 1)
  • The roots of 6x2 - x – 2 =0 are the values of x for which (3x – 2)(2x + 1) = 0. Therefore, 3x – 2 = or 2x + 1 = ,
  • Therefore, x = or x =
  • Therefore, the roots of 6x2 - x – 2 = 0 are 23 or −12
  • We verify the roots, by checking that 23 and −12 satisfy 6x2 - x – 2 =0.

Find the roots of the quadratic equation 3x2-26x +2 =0

Find the roots

  • We have 3x2-26x +2 split the term -26x as 3x2- - +2
  • Divide the eq into two part first term 3x2-6x written as √3x(√3x-) and second term -6x + 2 written as -2(3x-)
  • So common terms written as
  • So, the roots of the equation are the values of x for which (3x-2) (3x-2) =
  • Now, (3x-2) = 0 for x =
  • So, this root is repeated twice, one for each repeated factor Therefore, the roots of 3x2-26x+2 = 0 are ,

Find the dimensions of the prayer hall discussed in introduction part

Find the dimensions

  • In introduction part we found that if the breadth of the hall is m
  • then x satisfies the equation 2x2+ x – 300 =
  • Applying the factorisation method, we write this equation as 2x2- x + x – 300 = 0
  • Split the terms as 2x (x – ) + 25 (x – ) = 0
  • Therefore, (x – 12)(2x + 25) = 0
  • So, the roots of the given equation are x = or x = . Since x is the breadth of the hall, it cannot be negative.
  • Thus, the breadth of the hall is m. Its length = 2x + 1 = m.