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2 years ago

The methods of solving quadratic equations

1 answer:

8 0

There are three main ways of solving quadratic equations. The quadratic formula, factoring, and completing the square.

[1] using the quadratic formula

When the equation is in the form of ax² + bx + c = 0.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Read more about the quadratic formula here:

brainly.com/question/27750885

[2] factoring

Read more about how to solve with factoring here:

brainly.com/question/25829061

[3] completing the square

To start out with completing the square, make sure your constant is isolated on one side. The equation will be as ax² + bx = -c

Read more about completing the square here:

brainly.com/question/4822356

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Hey guys im new here please solve this for me with steps! ill mark as the best answer

Vinil7 [7]

Answer:

The factors of $2(x+y)^2-9(x+y)-5$ is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=> $2(x+y)^2-9(x+y)-5$

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then $2(x+y)^2-9(x+y)-5$ can be written as $2k^2-9k-5$

Now by using quadratic formula

k = $\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

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c= -5

Substituting the values, we get

k = $\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k = $\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k = $\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k = $\frac{-(-9) \pm \sqrt{(121)}}{4}$

k = $\frac{-(-9) \pm 11}}{4}$

k= $\frac{ 9 \pm 11}{4}$

k = $\frac{20}{4}$ k = $\frac{-2}{4}$

$k_1 =5$ $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

$(k-5)(2k+1)$

Substituting k = x+y

$((x+y)-5)(2(x+y+1)$

$((x+y)-5)(2x+2y+1)$

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