Solve equation by factoring.

Mathematics

1 answer:

3241004551 [841]3 years ago

8 0

Hi ;-)

$x^2+6x-10=30 \ \ /-30\\\\x^2+6x-40=0\\\\a=1, \ b=6, \ c=-40\\\\\Delta=b^2-4ac=6^2-4\cdot1\cdot(-40)=36+160=196\\\\\sqrt{\Delta}=\sqrt{196}=14\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-14}{2}=\boxed{-10}\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+14}{2}=\boxed4$

You might be interested in

Hi please help me choose the correct answer I will Mark you brainlst

Yuri [45]

Answer:

it's d

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

If a = (3-2√2) find the value of a4+ 1/a4

FinnZ [79.3K]

Answer:

1154

Step-by-step explanation:

Perhaps the easiest way to evaluate this numerical expression is to let a calculator do it. Alternatively, we can compute the value from a +1/a.

<h3>expression</h3>

Consider the square of a +1/a:

(a +1/a)^2 = a^2 +2(a)(1/a) +1/a^2 = (a^2 +1/a^2) +2

This means ...

a^2 +1/a^2 = (a +1/a)^2 -2

Similarly, using a^2 for 'a' in the above, we have ...

a^4 +1/a^4 = (a^2 +1/a^2)^2 -2

<h3>numerical value</h3>

The value of a +1/a is ...

$a+\dfrac{1}{a}=3-2\sqrt{2}+\dfrac{1}{3-2\sqrt{2}}=(3-2\sqrt{2})+\dfrac{3+2\sqrt{2}}{3^2-(2\sqrt{2})^2}\\\\=(3-2\sqrt{2})+(3+2\sqrt{2})\\\\a+\dfrac{1}{a}=6$