Question

Factorisation of middle terms:10_3x_x^2​

Answer 1

Given equation :

• 10 - 3x - x²

To Find :

• Values of x using middle term factorization.

Solution :

To solve any quadratic equation, it should first be brought to its general form.

General form :

• ax² + bx + c = 0

We can clearly see that the given equation is not in general form.

General form of the given equation :

• x² - 3x - 10 = 0

Now, let's find the zeroes of the equation.

➣ $\mathtt{x^2\:-3x-10=0}$

➣ $\mathtt{x^2\:-5x+2x-10=0}$

➣ $\mathtt{x(x-5)\:+2(x-5) =0}$

➣ $\mathtt{(x-5)\:\:(x+2)=0}$

➣ $\mathtt{x-5=0\:\:or\:\:x+2=0}$

➣ $\mathtt{x=5\:\:or\:\:x=-2}$

$\large{\boxed{\rm{\red{Roots\:of\:the\:equation\:are\::5,-2}}}}$

Answer 2

$\huge\boxed{\bf{Question :-}}$

Factorisation of middle term :- 10 x – 3 x – x²

$\huge {\boxed{\bf{Solution :-}}}$

$\bf \red{We \: have \: to \: find,}$

$\star \rm \: By \: using \: \red{Middle \: Term \: Factorisation}, \: we \: have \\ \rm to \: find \: the \: value \: of \: \red{x}.$

$\boxed{\dag \: \bf \green{ Remark}}$

For solving any type of quadratic equation, we have to brought it on it's Standard form. But we can see the given equation is not in standard form.

$\bf \red{Standard \: Form \: of \: any \: quadratic \: equation : - }$

$\boxed{ \longrightarrow \: \: \: \: \: \: \: \: \rm \: ax {}^{2} + bx + c = 0 \: }$

$\bf \red{Standard \: Form \: of \: the \: given \: Quadratic \: Equation : - }$

$\boxed{ \longrightarrow \: \: \: \: \: \: \: \rm \: x {}^{2} - 3x - 10 = 0 \: }$

Now, we have to find the zeroes of the given quadratic equation :-

$\rm \: We \: will \: find \: the \: zeroes \: by \: the \: form \: given \: form :-\\ \longrightarrow \: \bf \red{ax {}^{2} + bx + c = 0 \: }$

$\rm \: Let's \: find,$

$\boxed {\rightarrow \: \: \: \: \: \rm \: x {}^{2} - 3x - 10 = 0 \: }$

$\boxed{\rightarrow \: \: \: \: \: \rm \: x {}^{2} - 5x + 2x - 10 = 0 \: }$

$\boxed{ \rightarrow \: \: \: \: \: \rm \: x(x - 5) + 2(x - 5) = 0 \: }$

$\boxed{\rightarrow \: \: \: \: \: \rm \: (x - 5) \: (x + 2) = 0 \: }$

$\boxed{ \rightarrow \: \: \: \: \: \rm \: x - 5 = 0 \: \red{or} \: x + 2 = 0 \: }$

$\boxed{ \rightarrow \: \: \: \: \: \rm \: x = 5 \: \red{or} \: x = -2 \: }$

$\rm \: Hence, \: the \: roots \: of \: the \: given \: equation \:are \: \: \red{5 , \: - 2}$