Question

solve the following quadratic equations by factorization​

Answer 1

Given Question :-

Solve the following quadratic equations by factorization method :-

$\rm \: (a). \: \: 16x - \dfrac{10}{x} = 27$

$\rm \: (b). \: \: {3x}^{2} - 14x - 5 = 0$

$\rm \: (c). \: \: {5x}^{2} - 3x - 2 = 0$

$\large\underline{\sf{Solution-a}}$

Given quadratic equation is

$\rm \: 16x - \dfrac{10}{x} = 27$

$\rm \: \dfrac{ {16x}^{2} - 10}{x} = 27$

$\rm \: {16x}^{2} - 10 = 27x$

$\rm \: {16x}^{2} - 10 - 27x = 0$

$\rm \: {16x}^{2} - 27x - 10 = 0$

Using splitting of middle terms, we get

$\rm \: {16x}^{2} - 32x + 5x - 10 = 0$

$\rm \: 16x(x - 2) + 5(x - 2) = 0$

$\rm \: (x - 2)(16x + 5) = 0$

$\rm\implies \:x = 2 \: \: or \: \: x = - \dfrac{5}{16}$

$\large\underline{\sf{Solution-b}}$

Given quadratic equation is

$\rm \: {3x}^{2} - 14x - 5 = 0$

$\rm \: {3x}^{2} - 15x + x - 5 = 0$

$\rm \: 3x(x - 5) + 1(x - 5) = 0$

$\rm \: (x - 5)(3x + 1) = 0$

$\rm\implies \:x = 5 \: \: or \: \: x = - \dfrac{1}{3}$

$\large\underline{\sf{Solution-c}}$

Given quadratic equation is

$\rm \: {5x}^{2} - 3x - 2 = 0$

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

$\rm \: 5x(x - 1) + 2(x - 1) = 0$

$\rm \: (x - 1)(5x + 2) = 0$

$\rm\implies \:x = 1 \: \: or \: \: x = - \dfrac{2}{5}$

$\rule{190pt}{2pt}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

Question :-

• Solve the Quadratic Equations :-

$\bull \: \sf{16x - \frac{10}{x} = 27 }$

$\bull \: \sf{3 {x}^{2} - 14x - 5 = 0 }$

$\bull \: \sf{5 {x}^{2} - 3x - 2 = 0 }$

_____________________

Answer :-

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• Solving First Equation :-

$\dashrightarrow \: \sf{16x - \frac{10}{x} = 27 }$

$\dashrightarrow \: \sf{ \frac{16 {x}^{2} - 10 }{x} = 27 }$

$\dashrightarrow \: \sf{16x {}^{2} - 10 = 27x}$

$\dashrightarrow \: \sf{16 {x}^{2} - 10 - 27x = 0 }$

$\dashrightarrow \: \sf{16 {x}^{2} - 27x - 10 = 0 }$

$\dashrightarrow \: \sf{16 {x}^{2} - 32x + 5x - 10 = 0 }$

$\dashrightarrow \: \sf{16x \: (x−2)+5 \: (x−2)=0}$

$\dashrightarrow \: \sf{(x−2) \: (16x+5)=0}$

$\therefore \: \sf{x = 2 \: \: \: or \: \: \:x = \frac{5}{16} }$

⠀⠀⠀⠀⠀⠀⠀

• Solving Second Equation :-

$\dashrightarrow \: \sf{3 {x}^{2} - 14x - 5 = 0 }$

$\dashrightarrow \: \sf{3 {x}^{2} - 15x + x - 5 = 0}$

$\dashrightarrow \: \sf{3x \: (x - 5) + 1 \: (x - 5) = 0}$

$\dashrightarrow \: \sf{(x−5) \: (3x+1)=0}$

$\therefore \: \sf{x = 5 \: \: \: or \: \: \:x = - \frac{1}{3} }$

⠀⠀⠀⠀⠀⠀⠀

• Solving Third Equation :-

$\dashrightarrow \: \sf{5 {x}^{2} - 3x - 2 = 0 }$

$\dashrightarrow \: \sf{5 {x}^{2} - 5x+2x - 2=0 }$

$\dashrightarrow \: \sf{5x \: (x - 1) + 2 \: (x - 1) = 0}$

$\dashrightarrow \: \sf{(x - 1) \: (5x + 2) = 0}$

$\therefore \: \sf{x = 1 \: \: \: or \: \: \: x = - \frac{2}{5} }$

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