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Solve -

i) x^2 + 5x - 6 = 0

ii) 3x^2 + 9x + 6 = 0

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Answer 1

Given : The Quadratic Equations :

1. x² + 5x - 6 = 0
2. 3x² + 9x + 6 = 0

Need To Solve : The Given Quadratic Equations .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\Large{\underline {\underline {( 1)}}}$

⠀⠀⠀⠀⠀Quadratic Equation : $\qquad\:\:\bigg\lgroup \sf{ x^2 + 5x - 6 = 0 }\bigg\rgroup$

$\qquad \longmapsto \sf x^2 + 5x - 6 = 0$

$\sf\dag \:\:\:By\:Using \:Sum-Product \: Pattern\::$

$\qquad \longmapsto \sf x^2 + 6x - x - 6 = 0$

$\sf\dag \:\:\: Now,\:Finding \:out\:common \:terms\::$

$\qquad \longmapsto \sf x( x + 6) -1 (x + 6) = 0$

$\sf\dag \:\:\: Now,\:Rewrite \:in\:Factored \:terms\::$

$\qquad \longmapsto \sf (x - 1) (x + 6) = 0$

$\qquad \longmapsto \frak{\underline{\purple{\:x =\:\:1 \:\:or\:\:-6\:\: }} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:value \:of\:x \:is\:\bf{1 \:\:or\:\:-6\:\:}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\Large{\underline {\underline {(2)}}}$

⠀⠀⠀⠀⠀Quadratic Equation : $\qquad \:\:\bigg\lgroup \sf{ 3x^2 + 9x + 6 = 0 }\bigg\rgroup$

$\qquad \longmapsto \sf 3x^2 + 9x + 6 = 0$

$\dag\sf\:\:\:Now,\:By\:taking\:3\:\:as\:common\:in\:each\:term\::$

$\qquad \longmapsto \sf x^2 + 3x + 2 = 0$

$\sf\dag \:\:\:By\:Using \:Sum-Product \: Pattern\::$

$\qquad \longmapsto \sf x^2 + 2x + x + 2 = 0$

$\sf\dag \:\:\: Now,\:Finding \:out\:common \:terms\::$

$\qquad \longmapsto \sf x( x + 2) +1 (x + 2) = 0$

$\sf \dag\:\:\: Now,\:Rewrite \:in\:Factored \:terms\::$

$\qquad \longmapsto \sf (x + 1) (x + 2) = 0$

$\qquad \longmapsto \frak{\underline{\purple{\:x =\:\:-1 \:\:or\:\:-2\:\: }} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:value \:of\:x \:is\:\bf{-1 \:\:or\:\:-2\:\:}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Answer 2

★ Concept :-

Here the concept of Splitting the Middle Term has been used. We see that we are given two equations to be solved. These are quadratic equations which means that there will be two values of the variable. The values of the variables are called as the roots of the equation. Firstly we can make the middle term split in the form where it is dependent on other terms. Then we can take terms in common and then by grouping we can find the answers.

Let's do it !!

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★ Solution :-

i.) x² + 5x - 6 = 0

Here we see that,

• Product of coefficient of x² and constant term = 1 × 6 = 6

• Difference of the constant term and the coefficient of x² = 6 - 1 = 5

Now the given equation can be written as,

>> x² + 6x - 1x - 6 = 0

(Here we simply applied the correc sign qnd splitted the middle term in two terms)

Now let's take the common terms

>> x(x + 6) - 1(x + 6) = 0

On grouping we get,

>> (x - 1)(x + 6) = 0

Here either (x - 1) = 0 or (x + 6) = 0 since the product is zero so anyóne of the terms should be zero.

>> (x - 1) = 0 or (x + 6) = 0

>> x - 1 = 0 or x + 6 = 0

>> x = 1 or x = -6

Here we got two solutions of the equations. This means thet are in correct form.

$\;\underline{\boxed{\tt{Required\;\:value\;\:of\;\:x\;=\;\bf{\purple{1,\;-6}}}}}$

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ii.) 3x² + 9x + 6 = 0

Here we see that,

• Product of coefficient of x² and constant term = 3 × 6 = 18

• Sum of coefficient of x² ans constant term = 3 + 6 = 9

So the given equation can be written as,

>> 3x² + 6x + 3x + 6 = 0

On taking the common terms, we get

>> 3x(x + 2) + 3(x + 2) = 0

On grouping the common terms, we get

>> (3x + 3)(x + 2) = 0

Here either (3x + 3) = 0 or (x + 2) = 0. So,

>> 3x + 3 = 0 or x + 2 = 0

>> 3x = -3 or x = -2

>> x = -3/3 or x = -2

>> x = -1 or x = -2

These are the two roots of the equation.

$\;\underline{\boxed{\tt{Required\;\:value\;\:of\;\:x\;=\;\bf{\green{-1,\;-2}}}}}$

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★ More to know :-

• When the equation is too hard to use the method of splitting the middle term to find the roots. Then we use the Quadratic Formula or Sridharacharya Formula. By this formula, the value of x is given as ::

$\;\;\tt{\rightarrow\;\;x\;=\;\dfrac{-b\:\pm\:\sqrt{b^{2}\:-\:4ac}}{2a}}$

• a is the coefficient of x²

• b is the coefficient of x

• c is the constant term

This formula is applicable for the quadratic equations of the form ax² + bx + c