Question

factorise 2y square + 3y + 1​

Answer 1

Given :

$2y {}^{2} + 3y + 1 = 0$

What to find out=Roots of the equation?

Solution :-

Factorise by splitting middle term

$2 {y}^{2} + 3y + 1 = 0 \\ 2 {y}^{2} + 2y + y + 1 = 0 \\ 2y(y + 1) + 1(y + 1) = 0 \\ (y + 1)(2y + 1) = 0$

Now split it into possible cases

$(y + 1) = 0 \\ (2y + 1) = 0$

Hence,

$y_1 = - 1 \\ y_2 = \frac{ - 1}{2}$

Extra information :-

• A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.

Answer 2

Answer:

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QUESTION :

Factorise 2y² + 3y + 1 = 0

SOLUTION :

➡ 2y² + 3y + 1 = 0

➡ 2y² + 2y + y + 1 = 0

➡ 2y(y + 1) + 1(y + 1) = 0

➡ (2y + 1) (y + 1) = 0

➡ 2y + 1 = 0 ; y + 1 = 0

➡ 2y = 0 - 1 ; y = 0 - 1

➡ 2y = -1 ; y = -1

➡ y = -1/2 ; y = -1

Therefore, y = -1/2 & -1

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Step-by-step explanation:

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