Question

factorisation : 7x^2 + 5√7x + 4​

Answer 1

$\Large{\underbrace{\underline{\sf{Understanding\;the\; Concept}}}}$

Here we have given a quadratic equation and we have to factorise it to obtain the roots of the equation.

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We can factorise quadratic equations by 2 methods:

First- middle term splitting.

2nd- quadratic formula.

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So let's factorise it by middle term splitting!

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Given Quadratic Equation:

7x²+5√7x+4=0

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Middle term splitting

In middle term splitting method we have to split the middle term ( or coefficient of x) such that the sum of splitted term is equal to middle term and their product will be equal to the product of coefficient of x² and constant term.

7x²+5√7x+4=0

7x²+4√7x+√7x+4=0

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Now let's expand it more for better understanding:

√7√7x²+4√7x+√7+4=0

Now take common values aside:

√7x(√7x+4)+1(√7x+4)=0

(√7x+1)((√7x+4)=0

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First root:

√7x+1=0

√7x=-1

$\boxed{\sf{x=\dfrac{1}{\sqrt7}}}$

Now rationalise it:

$\sf{ x=\dfrac{1\times \sqrt7}{\sqrt7\times\sqrt7}}$

$\boxed{\sf{x=\dfrac{\sqrt7}{7}}}$

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Second root:

√7x+4=0

√7x=-4

$\boxed{\sf{ x=\dfrac{-4}{\sqrt7}}}$

Now rationalise it:

$\sf{ x=\dfrac{-4\times\sqrt7}{\sqrt7\times\sqrt7}}$

$\boxed{\sf{x=\dfrac{-4\sqrt7}{7}}}$

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So the roots are:-

$\sf{\dfrac{\sqrt7}{7}\;and\; \dfrac{-4\sqrt7}{7}}$

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