Class 10 Solution of a Quadratic Equation by Factorisation

Please Wait... While Loading Full Video

Class 10 Chapter 4 - QUADRATIC EQUATIONS

Solution of a Quadratic Equation by Factorisation

Topic to be coverd

`♦` Solution of a Quadratic Equation by Factorisation

`♦` Solution of a Quadratic Equation by Factorisation

Solution of a Quadratic Equation by Factorisation

`color{green} ✍️` In general, a real number `α` is called `color(green)("a root of the quadratic equation")` `color(blue)(ax^2 + bx + c = 0, a ≠ 0)`

if `color(red)(a α^2 + bα + c = 0.)` We also say that `x = α` is a solution of the quadratic equation, or that `color(blue)("α satisfies the quadratic equation.")`

`color(green)(Note )` that the zeroes of the quadratic polynomial `ax^2 + bx + c` and the roots of the quadratic equation `ax^2 + bx + c = 0` are the same.

`color(red)(=>"Consider the quadratic equation")` `color(red)(2x^2 – 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 `2x^2 – 3x + 1 = 0.`

This also means that `1` is a zero of the quadratic polynomial `2x^2 – 3x + 1.`

You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes.

`color{green} ✍️` `color(green)("So, any quadratic equation can have atmost two roots.")`

`color{green} ✍️` You have learnt in Class IX, how to factorise quadratic polynomials by splitting their middle terms. We shall use this knowledge for finding the roots of a quadratic equation.

`color{green} ✍️` In general, a real number `α` is called `color(green)("a root of the quadratic equation")` `color(blue)(ax^2 + bx + c = 0, a ≠ 0)`

if `color(red)(a α^2 + bα + c = 0.)` We also say that `x = α` is a solution of the quadratic equation, or that `color(blue)("α satisfies the quadratic equation.")`

`color(green)(Note )` that the zeroes of the quadratic polynomial `ax^2 + bx + c` and the roots of the quadratic equation `ax^2 + bx + c = 0` are the same.

`color(red)(=>"Consider the quadratic equation")` `color(red)(2x^2 – 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 `2x^2 – 3x + 1 = 0.`

This also means that `1` is a zero of the quadratic polynomial `2x^2 – 3x + 1.`

You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes.

`color{green} ✍️` `color(green)("So, any quadratic equation can have atmost two roots.")`

`color{green} ✍️` You have learnt in Class IX, how to factorise quadratic polynomials by splitting their middle terms. We shall use this knowledge for finding the roots of a quadratic equation.

Q 3169767615

Find the roots of the equation `2x^2 – 5x + 3 = 0,` by factorisation.
Class 10 Chapter 4 Example 3

Solution:

Let us first split the middle term `– 5x` as `–2x –3x` [because `(–2x) × (–3x) = 6x^2 = (2x^2) × 3`].
So, `2x^2 – 5x + 3 = 2x^2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)`
Now, `2x^2 – 5x + 3 = 0` can be rewritten as `(2x – 3)(x – 1) = 0.`
So, the values of `x` for which `2x^2 – 5x + 3 = 0` are the same for which `(2x – 3)(x – 1) = 0,`
i.e., either `2x – 3 = 0` or `x – 1 = 0.`
Now, `2x – 3 = 0` gives `x = 3/2` and `x-1 = 0` gives `x = 1`.

so `x = 3/2` and `x = 1` are the solutions of the eqaution.

In other words , 1 and `3/2` are the roots of the equation `2x^2-5x+3 = 0`.

Verify that these are the roots of the given equation.

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

Q 3189767617

Find the roots of the quadratic equation `6x^2 – x – 2 = 0.`
Class 10 Chapter 4 Example 4

Solution:

We have
`6x^2 – x – 2 = 6x^2 + 3x – 4x – 2`
`= 3x (2x + 1) – 2 (2x + 1)`
`= (3x – 2)(2x + 1)`
The roots of `6x^2 – x – 2 = 0` are the values of x for which `(3x – 2)(2x + 1) = 0`
Therefore, `3x – 2 = 0` or `2x + 1 = 0,`

i.e. `x = 2/3` or `x = -1/2`

Therefore, the roots of `6x^2 – x – 2 = 0` are `2/3` and `-1/2`.

We verify the roots, by checking that `2/3` and `-1/2` satisfy `6x^2-x-2 = 0`

Q 3129867711

Find the roots of the quadratic equation `3x^2 - 2 sqrt6x + 2 = 0 .`
Class 10 Chapter 4 Example 5

Solution:

`3x^2 - 2 sqrt(6x) + 2 = 3x^2-sqrt6x - sqrt6x +2`

` = sqrt3x( sqrt3x-sqrt2) - sqrt2 ( sqrt3x-sqrt2)`

`= ( sqrt3x - sqrt2) ( sqrt3x-sqrt2)`

So, the roots of the equation are the values of x for which

`( sqrt3x-sqrt2) ( sqrt3x-sqrt2) = 0`

Now `sqrt3x-sqrt2 = 0` for `x = sqrt(2/3)`

So, this root is repeated twice, one for each repeated factor `sqrt3x-sqrt2`.

Therefore, the roots of `3x^2-2 sqrt6x+2 = 0` are `sqrt(2/3) , sqrt(2/3)`

Q 3189067817

Find the dimensions of the prayer hall discussed in a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall ?
Class 10 Chapter 4 Example 6

Solution:

Suppose the breadth of the hall is `x` metres. Then,
its length should be `(2x + 1)` metres. We can depict
this information pictorially as shown in Fig.
Now, area of the hall` = (2x + 1). x m^2 = (2x^2 + x) m^2`
So, `2x^2 + x = 300` (Given)
Therefore, `2x^2 + x – 300 = 0`

Applying the factorisation method, we write this equation as

`2x^2 – 24x + 25x – 300 = 0`
`2x (x – 12) + 25 (x – 12) = 0`
i.e.,` (x – 12)(2x + 25) = 0`
So, the roots of the given equation are `x = 12` or `x = – 12.5.` Since `x` is the breadth of the hall, it cannot be negative.
Thus, the breadth of the hall is `12 m`. Its length `= 2x + 1 = 25 m.`

SiteLock