X^2-6x-25=0. Solve by completing the square.

1 answer:

7 0

Hey there,
x^2 -6x+25=0
Per quadratic formula, roots of ax^2+bx+c are
x=[-b+sqrt(b^2-4ac)]/2a and x=[-b-sqrt(b^2-4ac)]/2a

In this example, a=1, b=-6, and c=25
So, x= [6+sqrt(36-100)]/2 and x= [6-sqrt(36-100)]/2
=> x = [6+sqrt(-64)]/2 and x = [6-sqrt(-64)]/2
=> x = 3+4i and x= 3-4i

Hope this helped you alot!!!!!!!!

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aivan3 [116]

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6 0

3 years ago

the product of two consecutive positive even integers is 14 more than their sum. Set up an equation that can

alexandr1967 [171]

Required equation to determine two consecutive even number is $x^{2}-16=0$ and required numbers are 4 and 6.

Solution:

Given that product of two consecutive positive integer is 14 more that their sum. Need to create the equation and solve it to get two numbers.

Let’s assume first even number be represented by x.

So second consecutive even number will be $x + 2$

Sum of two consecutive number = $(x) + (x + 2)$

$\Rightarrow$ Sum of two consecutive number = $2x + 2$

Product of two consecutive number = $(x) \times(x+2)=x^{2}+2 x$

Product of two consecutive number is 14 more that sum of two consecutive numbers, so if we subtract 14 from product, we will get sum.

$\begin{array}{l}{\Rightarrow\left(x^{2}+2 x\right)-14=2 x+2} \\\\ {=>x^{2}+2 x-2 x-14-2=0} \\\\ {\Rightarrow x^{2}-16=0} \\\\ {\Rightarrow x^{2}-4^{2}=0} \\\\ {\Rightarrow x^{2}=4^{2}} \\\\ {\Rightarrow x=4}\end{array}$