Math Help ASAP! Please help! Urgent!?
If you could help me out it would be greatly appreciated! My parents and I have been trying to figure these out all day, along with calling other relatives to see if they could figure it out. My teacher doesn't really explain things to you at all if you've been absent. So obviously she didn't give me the slightest clue on what to do once I got back from being out with the flu for a week 1) What values of a, b, and c would you use in the quadratic formula to solve: (note there are 2 equations) A: 3x² – 7x = 9? B: – 7x = 9 + x – 2x² 2) . Solve 3x²– 7x = 9. Estimate irrational solutions to the nearest tenth. 3) . Solve: For x² + 2x – 9 = 7 – x – 5x² A: Find the discriminant. B: Give the number of real solutions. C: Give the number of rational solutions.
Find the discriminant: 2x^2-6x+2=0?
using the formula b^2-4ac given a=2, b=-6, c=2 =(-6)^2-4(2)(2) =36-16 =20 answer is b)20
Find the discriminant of x^2 - 10x + 25?
The discriminant is 0 because x^2 - 10x + 25 = (x-5)^2. For the second one we have 5x -10 = x^2 -5x + 15 x^2 -10x + 25 = 0 So x = 5 is the only possibility , and if x = 5, y = 15. The two curves meet only at the point (5,15).
Algebra help ASAP!? Please I dont understand?
1. 2x² + 17x − 9 = 2x² + 18x − x − 9 = 2x (x + 9) − 1 (x + 9) = (x + 9) (2x − 1) B (2x − 1) ------------------------------ 2. 10x² + 11x + 3 b²−4ac = 121−4(10)(3) = 121−120 = 1 Since discriminant is a perfect square, trinomial is NOT prime Answer: FALSE ------------------------------ 3. 4x² + 24x + 11 = 4x² + 2x + 22x + 11 = 2x (2x + 1) + 11 (2x + 1) = (2x + 1) (2x + 11) A (2x + 1) ------------------------------ 4. 2x² − 8x + 5 b²−4ac = 64−4(2)(5) = 64−40 = 24 Since discriminant is a NOT perfect square, trinomial IS prime Answer: TRUE ------------------------------ 5. 9x² − 18xy + 5y² = 9x² − 3xy − 15xy + 5y² = 3x (3x − y) − 5y (3x − y) = (3x − y) (3x − 5y) C (3x − 5y) Mαthmφm
The line 4x+y+k is a tangent to the parabola y= 3x^2 + 8x -9. Find the value of k?
I am assuming that the given equation of the line is: 4x + y = k. First, put this line in slope-intercept form by solving for y to get: 4x + y = k ==> y = -4x + k. One way to find the value of k without Calculus to note that if y = -4x + k is tangent to y = 3x^2 + 8x - 9, then the line will only intersect y = 3x^2 + 8x - 9 once. By equating the values of x: 3x^2 + 8x - 9 = -4x + k ==> 3x^2 + 12x + (-k - 9) = 0. Then, only one intersection will occur when this equation only has one solution, which occurs when the discriminant is zero. The discriminant of ax^2 + bx + c = 0 is b^2 - 4ac, so the discriminant of this equation is: b^2 - 4ac = 12^2 - 4(3)(-k - 9) = 12k + 252. This equals zero when k = -21. Therefore, k = -21. I hope this helps!
Roots of quadratic equations! ( math geniuses i summon you .. =') please help !?
Determine the roots of each quadratic equation using a method of your choice. please show your work so i can understand how you went through it. a. v2-2v-24=0 (2) Answer: b.11-6x=0.8x2 ( to the power of two ) Answer: 3. The length of a rectangular footprint of a residence is 5 m more than its width. If the area of the rectangle is 104 m2, what are the dimensions of the rectangle? Answer: 4. Determine the values of c in the equation x2+8x+c=0 that will result in a. two different real roots b. one distinct real root c. no real roots Answer: guys please help with what you can !! and please try to show your work so i can go through it ..!
Please help me solve these questions pleaseee ???!!!!!!!?
= (a + √5)(3 - b√5) → you expand = 3a - ab√5 + 3√5 - 5b → you group = (3a - 5b) + √5(3 - ab) → you compare with: 26 + 11√5 → and you can see that: (1) : 3a - 5b = 26 (2) : 3 - ab = 11 From the equation (1): → a = (26 + 5b)/3 You substitute a by its value into the equation (2) 3 - ab = 11 ab = - 7 [(26 + 5b)/3] * b = - 7 b(26 + 5b) = - 21 5b² + 26b + 21 = 0 Polynomial like: ax² + bx + c, where: a = 5 b = 26 c = 21 Δ = b² - 4ac (discriminant) Δ = 26² - 4(5 * 21) = 676 - 420 = 256 = 16² x1 = (- b - √Δ) / 2a = (- 26 - 16) / (2 * 5) = - 21/5 x2 = (- b + √Δ) / 2a = (- 26 + 16) / (2 * 5) = - 1 First case: b = - 21/5 Recall: a = (26 + 5b)/3 a = [26 + 5(- 21/5)]/3 a = [26 - 21]/3 a = 5/3 Second case: b = - 1 Recall: a = (26 + 5b)/3 a = [26 + 5(- 1)]/3 a = [26 - 5]/3 a = 19/3 Solution for (a ; b) → (5/3 ; - 21/5) and (19/3 ; - 1) The length (L) of one side is (√3 + √2) The length (ℓ) of the other side is unknown The area of the rectangle is: (1 + √6) m² L * ℓ = (1 + √6) ℓ = (1 + √6)/L → but you know that: L = √3 + √2 ℓ = (1 + √6)/(√3 + √2) → then multiply by: (√3 - √2) ℓ = [(1 + √6)(√3 - √2)] / [(√3 + √2)(√3 - √2)] ℓ = [√3 - √2 + √18 - √12] / [(√3)² - √6 + √6 - (√2)²] ℓ = [√3 - √2 + √18 - √12] / (3 - 2) ℓ = √3 - √2 + √18 - √12 → you know that: √18 = √(9 * 2) = √9 * √2 = 3√2 ℓ = √3 - √2 + 3√2 - √12 → you know that: √12 = √(4 * 3) = √4 * √3 = 2√3 ℓ = √3 - √2 + 3√2 - 2√3 ℓ = 2√2 - √3
Solve 3x2 + 2x + 7 = 0. Round solutions to the nearest hundredth?
1. By quad. formula.. ax² + bx + c = 0 a = 3 b = 2 c = 7 Then.. x = (-b ± √(b² - 4ac))/2a = (-2 ± √(2² - 4*3*7))/(2*3) = (-2 ± √(4 - 84))/6 = (-2 ± √(-80))/6 Since √-80 is imaginary, there are no real solutions. 2. 3x² - x - 4 = 0 By factoring, we obtain: (3x - 4)(x + 1) = 0 Finallly, by zero-product property.. 3x - 4 = 0 and x + 1 = 0 x = {4/3, -1} I hope this helps!