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3 years ago

6

Factorise: a squared +8a -20

1 answer:

Vaselesa [24]3 years ago

8 0

Factor out the 4 in both equations
8a^2-20^2=(2^2 times a^2 times 2)-(2^2 times 5)
therefor it is also equal to
(2a)^2 times 2-(2^2 times 5)
we can force it into a difference of 2 perfect squares which is a^2-b^2=(a-b)(a+b)
(2a√2)^2-(2√5)^2=((2a√2)-(2√5))((2a√2)+(2√5))

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Y=.5x+4 <br> Y= -.25x^2+5 <br> Using the quadratic formula, what are the solutions?

Aleksandr-060686 [28]

Hi there!

$\large\boxed{x = -1 \pm \sqrt{5}}$

We can solve by setting both equations equal to each other:

.5x + 4 = -.25x² + 5

Move everything to one side:

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Multiply both sides by 4 to cancel out the decimals:

0 = -x² - 2x + 4

Factor out negative from both sides:

0 = x² + 2x - 4

Use the quadratic formula:

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

Plug in the corresponding terms:

$x = \frac{-2 \pm \sqrt{4-4(-4)(1)}}{2}$

Simplify:

$x = \frac{-2 \pm \sqrt{4 + 16}}{2}\\\\x = \frac{-2 \pm \sqrt{20}}{2}$

Simplify the radical to reduce:

$x = \frac{-2 \pm 2\sqrt{5}}{2}$

Divide all terms by 2:

$x = -1 \pm \sqrt{5}$

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