Factorisation : 7x^2 + 5√7x + 4

1.	Factorisation : 7x^2 + 5√7x + 4
Answer» $\Large{\underbrace{\underline{\sf{Understanding\;the\; Concept}}}}$ Here we have given a quadratic equation and we have to factorise it to obtain the ROOTS of the equation. $\rule{300}{2}$ We can factorise quadratic equations by 2 METHODS: First- middle term splitting. 2nd- quadratic formula. $\rule{300}{2}$ So let's factorise it by middle term splitting! $\rule{300}{2}$ Given Quadratic Equation: 7x²+5√7x+4=0 $\rule{300}{2}$ 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 $\rule{300}{2}$ 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 $\rule{300}{2}$ First root: √7x+1=0 √7x=-1 $\boxed{\sf{x=\dfrac{1}{\sqrt7}}}$ Now rationalise it: $\sf{ x=\dfrac{1\times \sqrt7}{\sqrt7\times\sqrt7}}$ $\boxed{\sf{x=\dfrac{\sqrt7}{7}}}$ $\rule{300}{2}$ SECOND root: √7x+4=0 √7x=-4 $\boxed{\sf{ x=\dfrac{-4}{\sqrt7}}}$ Now rationalise it: $\sf{ x=\dfrac{-4\times\sqrt7}{\sqrt7\times\sqrt7}}$ $\boxed{\sf{x=\dfrac{-4\sqrt7}{7}}}$ $\rule{300}{2}$ So the roots are:- $\sf{\dfrac{\sqrt7}{7}\;and\; \dfrac{-4\sqrt7}{7}}$ $\rule{300}{2}$

Answer»

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

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

$\rule{300}{2}$

We can factorise quadratic equations by 2 METHODS:

First- middle term splitting.

2nd- quadratic formula.

$\rule{300}{2}$

So let's factorise it by middle term splitting!

$\rule{300}{2}$

Given Quadratic Equation:

7x²+5√7x+4=0

$\rule{300}{2}$

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

$\rule{300}{2}$

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

$\rule{300}{2}$

First root:

√7x+1=0

√7x=-1

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

Now rationalise it:

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

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

$\rule{300}{2}$

SECOND root:

√7x+4=0

√7x=-4

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

Now rationalise it:

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

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

$\rule{300}{2}$

So the roots are:-

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

$\rule{300}{2}$

Factorisation : 7x^2 + 5√7x + 4

We can factorise quadratic equations by 2 METHODS: