Factorise the given quadratic equation x-2 nbsp-5x-6 -0-x-2-x-3-x-2-x-3-x-1-x-3- nbsp- nbsp-x-1-x-3- nbsp-

The correct option is A (x+2)(x+3) Comparing x^2 nbsp;+5x+6 =0 with nbsp;ax^2 nbsp;+ bx+ c = 0, we get nbsp;a = 1 , b = 5 and c = 6. Now, we need to fi ...

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Standard X

Mathematics

Factorisation

Factorise the...

Question

Factorise the given quadratic equation

$x^{2} + 5 x + 6 = 0.$

$(x + 2) (x + 3)$

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$(x - 2) (x - 3)$

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$(x + 1) (x + 3)$

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$(x - 1) (x - 3)$

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Solution

The correct option is A
$(x + 2) (x + 3)$

Comparing $x^{2} + 5 x + 6 = 0$ with $a x^{2} + b x + c = 0$ ,
we get a = 1 , b = 5 and c = 6.

Now, we need to find two numbers whose product is 6 and whose sum is 5.

Pairs of numbers whose product is 6 are

$(1, 6), (- 1, - 6), (2, 3)$ and $(- 2, - 3)$

Of these pairs , the pair that gives the sum as 5 is $(2, 3) .$

We can rewrite the given quadratic equation as

$x^{2} + 5 x + 6 = x^{2} + 2 x + 3 x + 6$
$= x (x + 2) + 3 (x + 2)$
$= (x + 2) (x + 3)$

Hence, the required answer is $(x + 2) (x + 3)$ .

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Similar questions

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(iii)

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(iv)

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(v)

2 x^{2} - \sqrt{3 x} + 9 = 0

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Q. Solve the following quadratic equations by factorization:

\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}

Check whether the following are quadratic equations :
(i) ${(x + 1)}^{2} = 2 (x - 3)$
(ii) $x^{2} - 2 x = (- 2) (3 - x)$
(iii) $(x - 2) (x + 1) = (x - 1) (x + 3)$
(iv) $(x - 3) (2 x + 1) = x (x + 5)$
(v) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
(vi) $x^{2} + 3 x + 1 = {(x - 2)}^{2}$
(vii) ${(x + 2)}^{3} = 2 x (x^{2} - 1)$
(viii) $x^{3} - 4 x^{2} - x + 1 = {(x - 2)}^{3}$

Solve the following quadratic equation by factorization.
$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$