bzoe8190
Answered

Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

please guys please pllease

Please Guys Please Pllease class=

Sagot :

Answer:

x= -3.936 and -0.064

Step-by-step explanation:

I simplified the equation to 40u^2 +16u + 1. Then I just graphed it and found that the zeroes were -3.936 and -0.064.

DᴀʀᴋPᴀʀᴀᴅᴏx

[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

Let's calculate its discriminant :

[tex]\qquad \sf  \dashrightarrow \: 4 {u}^{2} + 16u + 41 = 40[/tex]

[tex]\qquad \sf  \dashrightarrow \: 4 {u}^{2} + 16u + 41 - 40 = 0[/tex]

[tex]\qquad \sf  \dashrightarrow \: 4 {u}^{2} + 16u + 1 = 0[/tex]

Here, if we equate it with general equation,

  • a = 4

  • b = 16

  • c = 1

[tex]\qquad \sf  \dashrightarrow \: disciminant = {b}^{2} - 4ac[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = (16) {}^{2} - (4 \times 4 \times 1) [/tex]

[tex]\qquad \sf  \dashrightarrow \: d = (16) {}^{2} - (16) [/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 16(16 - 1)[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 16(15)[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 240[/tex]

Now, since discriminant is positive ; it has two real roots ~

The roots are :

[tex]\qquad \sf  \dashrightarrow \: u = \dfrac{ - b \pm \sqrt{ d } }{2a} [/tex]

[tex]\qquad \sf  \dashrightarrow \: u = \dfrac{ - 16\pm \sqrt{ 240 } }{2 \times 4} [/tex]

[tex]\qquad \sf  \dashrightarrow \: u = \dfrac{ - 16\pm 4\sqrt{ 15 } }{8} [/tex]

[tex]\qquad \sf  \dashrightarrow \: u = \dfrac{ 4(- 4\pm \sqrt{ 15 }) }{8} [/tex]

[tex]\qquad \sf  \dashrightarrow \: u = \dfrac{ - 4\pm \sqrt{ 15 } }{2} [/tex]

So, the required roots are :

[tex]\qquad \sf  \dashrightarrow \: u = \dfrac{ - 4 - \sqrt{ 15 } }{2} \: \: and \: \: \dfrac{ - 4 + \sqrt{15} }{2} [/tex]

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.