What are the approximate solutions of 2x2 - 7x = 3, rounded to the nearest hundredth? No real solutions x ≈ -0.77 and x ≈ 7.77 x ≈ -0.39 and x ≈ 3.89 x ≈ -3.89 and x ≈ 0.39
@satellite73
@iGreen
I think they want you to use the quadratic formula..
tbh i am not sure.... I got this big exam and im a little lost on some of these questions
Okay, hold on.
ok, and thank you man!
First we re-arrange the equation: \(2x^2 - 7x = 3\) Subtract -3 to both sides: \(2x^2 - 7x - 3 = 0\) Our equation is in the form of \(ax^2 + bx = c\) Now we can plug it into the quadratic formula. Formula: \(x =\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) \(a = 2\) \(b = -7\) \(c = -3\) \(x =\dfrac{-(-7) \pm \sqrt{-7^2 - 4(2)(-3)}}{2(2)}\) Simplify: \(x =\dfrac{7\pm \sqrt{-7^2 - 4(2)(-3)}}{4}\) Simplify Exponent: \(x =\dfrac{7\pm \sqrt{49 - 4(2)(-3)}}{4}\) Multiply: \(x =\dfrac{7\pm \sqrt{49 - 8(-3)}}{4}\) Multiply: \(x =\dfrac{7\pm \sqrt{49 + 24}}{4}\) Add: \(x =\dfrac{7\pm \sqrt{73}}{4}\) Now we split it into two equations. \(x =\dfrac{7+ \sqrt{73}}{4}\) and \(x =\dfrac{7- \sqrt{73}}{4}\) -------------------------- \(x =\dfrac{7+ \sqrt{73}}{4}\) Simplify the Square Root: \(x =\dfrac{7+ 8.54}{4}\) Add: \(x =\dfrac{15.54}{4}\) Divide: \(x \approx3.89\) -------------------------- \(x =\dfrac{7- \sqrt{73}}{4}\) Simplify the Square Root: \(x =\dfrac{7- 8.54}{4}\) Subtract: \(x =\dfrac{-1.54}{4}\) Can you divide that? @SouthernRebel101
am i dividing the last thing? the -1.54 over 4?
@iGreen
i am a little confused lol i divided and got -0.385
Yep, that's it, that rounds to -0.39 So: \(x \approx 3.89\) and \(x \approx -0.39\)
So what's your answer choice gonna be?
oh ok um.. C?
i mean D
wait no it would be C right
Yep^^ C is right.
xD i am all over
thank you, ill ask for more help when i need it thanky ou so much
\(\bf x ≈ -0.39~and~x ≈ 3.89\)
No problem. Just tag me when you do :P
oh i do have a question ill tagg you
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