Question

Solve the following quadratic equations
by factorisation.
2y² + 274+13=0

​

Answer 1

Good question!!

First what is quadratic equations?

Ans:-The polynomial having power or degree 2 is called quadratic polynomial.

when it is equated by 0 it forms quadratic equations.

It's general formula is :-

$a {x}^{2} + bx + c$

Where a,b,c are Coefficents and A not equal to 0.

Given:-

$2 {y}^{2} + 274y + 13$

$d = {b }^{2} - 4ac$

$d = {274}^{2} - 4 \times 2 \times 13$

$d = 75076 - 104$

$d = 74972$

Now using :-

X=-b+√d

2a

Answer 2

Correct Question:

Solve the following quadratic equation by factorisation:

$2{y}^{2} + 27y + 13 = 0$

Answer:

$The \: roots \: of \: the \: given \: quadratic \: equation \: are \ \: \\ </p><p></p><p>\boxed{y = - 13} \: \: OR \: \: \boxed{y = \frac{ - 1}{2} }$

Step-by-step-explanation:

$The \: given \: quadratic \: equation \: is \\ 2 {y}^{2} + 27y + 13 = 0 \\ \\ ⟹ \: 2 {y}^{2} + 26y + y + 13 = 0 \\ \\ ⟹ \: 2y(y + 13) \: + 1(y + 13) = 0 \\ \\ ⟹ \: (y + 13) \: (2y + 1) = 0 \\ \\ ⟹ \: y + 13 = 0 \: \: \: OR \: \: 2y + 1 = 0 \\ \\ ⟹ \: y = - 13 \: \: OR \: \: 2y = - 1 \\ \\ ⟹ \: y = - 13 \: \: \: OR \: \: \: y = \frac{ - 1}{2} \\ \\ </p><p></p><p>The \: roots \: of \: the \: given \: quadratic \: equation \: are \ \: \\ </p><p></p><p>\boxed{y = - 13} \: \: OR \: \: \boxed{y = \frac{ - 1}{2} } \\ \\ Verification: \: \\ \\The \: first \: root \: of \: the \: equation \: is \: y = - 13 \\ \\ ∴ \: LHS = \:2 {y}^{2} + 27y + 13 = 0 \\ \\ ⟹ \: 2 \times ( { - 13)}^{2} + 27 \times ( - 13) + 13 \\ \\ ⟹ \: 2 \times 169 - 351 + 13 \\ \\ ⟹ \: 338 - 338 \\ \\ ⟹ \: 0 \\ \\ RHS = 0 \: \\ \\ ∴ \boxed{LHS = RHS} \: \\ \\ The \: second \: root \: of \: the \: equation \: is \: y = \frac{ - 1}{2} \\ \\ ∴ \: LHS = \:2 {y}^{2} + 27y + 13 = 0 \: \ \\ \\ ⟹ \: 2 \times ( { \frac{ - 1}{2} )}^{2} + 27 \times ( \frac{ - 1}{2} ) + 13 \\ \\ ⟹ \: 2 \times \frac{1}{4} - \frac{ 27}{2} + 13 \\ \\ ⟹ \: \frac{2}{4} - \frac{27}{2} + 13 \\ \\ ⟹ \: \frac{1}{2} - \frac{27}{2} + 13 \\ \\ ⟹ \: \frac{1 - 27}{2} + 13 \\ \\ ⟹ \: \frac{ - 26}{2} + 13 \\ \\ ⟹ \: - 13 + 13 \\ \\ ⟹ \: 0 \\ \\ RHS = 0 \: \\ \\ ∴ \: \boxed{LHS = RHS} \ \\ \\ Hence \: verified \: ! \:$

Additional Information:

1. Quadratic Equation :

An equation having a degree '2' is called quadratic equation.

The general form of quadratic equation is

ax² + bx + c = 0

Where, a, b, c are real numbers and a ≠ 0.

2. Roots of Quadratic Equation:

The roots means nothing but the value of the variable given in the equation.

3. Methods of solving quadratic equation:

There are mainly three methods to solve or find the roots of the quadratic equation.

A) Factorization method

B) Completing square method

C) Formula method

4. Solution of Quadratic Equation by Factorization:

1. Write the given equation in the form ${ax^{2} + bx + c = 0}$

2. Find the two linear factors of the ${LHS}$ of the equation.

3. Equate each of those linear factor to zero.

4. Solve each equation obtained in 3 and write the roots of the given quadratic equation.