F x x 2 + 4kx + 3+11k where k is a constant

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August undefined, 2024

WebMar 5, 2015 · the question is : By considering the discriminant, or otherwise, find the range of values of 'k' that gives the equation 2 distinct roots. 3 x 2 + k x + 2 = 0 this is what i have done so far: b 2 − 4 a c > 0 Note - since there are two distinct real roots k 2 − 24 > 0 considering k 2 − 24 = 0 k = + − 24 k = 2 ∗ 6 or k = − 2 ∗ 6 WebSep 16, 2024 · fx=x2+4kx+3+11k, where k is a constant.. a Express fx in the form x+p2+q , where p and q are constants to be found in terms of k. 3 Given that the equation fx=0 has no real roots, b find the set of possible values of k. 4 Given that k=1, c sketch the graph of y=fx , showing the coordinates of any point at which the graph crosses a coordinate …

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WebF(x)=x^2+4kx+(3+11k) - f(x) = x2 + 4kx + (3 + 11k), where k is a constant. (a) Express f(x) in the form (x + p)2 + q, where p and q are constants to be found Math Practice WebMaths Genie - Free Online GCSE and A Level Maths Revision ezekiel jasone osoro

F(x)=x^2+4kx+(3+11k) - Math Practice

WebSep 17, 2016 · Sketching a graph using completing the square. For the question $f (X) = X^2 + 4KX + (3+11K)$, where $K$ is a constant. given that $K = 1$, Sketch the graph … WebJul 25, 2015 · 2 Answers. Sorted by: 1. If you don't want to incorporate your prior knowledge to the problem, the quadratic equation has equal roots when the discriminant is equal to zero. ( k − 1) 2 + 4 ( 1) ( 1 + 2 k) = k 2 + 6 k + 5 = ( k + 1) ( k + 5) = 0 k ∈ { − 5, − 1 } Share. Web0:00 / 2:45 Complete the square to find the vertex: f (x)=-2x^2+8x+3 Mark Dwyer 3.99K subscribers Subscribe 2.2K views 8 years ago Completing the Square This video … ezekiel jackson moves

How do you find the intercepts, vertex and graph f(x)=-2x^2+8x-3 ...

F x x 2 + 4kx + 3+11k where k is a constant

WebSilver 4: 9/15 4 6. The equation kx2 + 4x + (5 – k) = 0, where k is a constant, has 2 different real solutions for x. (a) Show that k satisfies k2 – 5k + 4 > 0. (3) (b) Hence find the set of possible values of k.(4) January 2009 7. A sequence ...a a a 1 2 3 , , , is defined by 1 a k, n 1 a a n 3 5 , n 1, where k is a positive integer. (a) Write down an expression for a2 … WebFirstly, you have to find the constant of k by using the discriminant. = (4k) 2 – (4 x 1 x (3+11k)) = 16k 2 – 44k – 12 Then use the quadratic formula to find the two values of k. For ease, I find the two values of k as 3 and-0.25. Secondly, replace your values of k into the original function which should give you a simpler quadratic equation. Then you will be …

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WebFeb 21, 2024 · x2 −4x + 3 2 = 0 [divide all the terms by −2] x2 −4x = − 3 2 [Take the constant term to the right] x2 −4x + 4 = − 3 2 + 4 (divide the coefficient of x square it and add to both sides] x2 −4x + 4 = − 3 2 + 4 = −3 + 8 2 = 5 2 (x −2)2 = − 5 2 x − 2 = ± √5 2 = ± 1.58 x = 1.58 + 2 = 3.58 x = − 1.58 +2 = 0.42 X intercept (3.58,0); (0.42,0) Y-Intercept WebQ^ (7). f (x)=x^ (2)+4kx+ (3+11k), where k is a constant Express f (x) in the form (x+p)^ (2)+q, where p and q are constants to be found in terms of k Expert Answer 1st step All steps Answer only Step 1/1 To express f (x) in the form ( x + p) 2 + q, we can use the technique of completing the square.

Webcompleting the square question. f (x) = x 2 + 4kx + (3 + 11k) where k is a constant. a) Express f (x) in the form (x+p) 2 + q where p and q are constants to be found in terms of … Webfx=x2+4kx+3+11k, where k is a constant.. a Express. f(x) = x2 + 4kx + (3 + 11k), where k is a constant. (a) Express f(x) in the form (x + p)2 + q, where p and q are constants to be found in terms of k. (3).

Webfx=x2+4kx+3+11k , where k is a constant. Given mar equation fx=0 has no real root find the ser or porrible values or k. WebQ^ (7). f (x)=x^ (2)+4kx+ (3+11k), where k is a constant Express f (x) in the form (x+p)^ (2)+q, where p and q are constants to be found in terms of k Expert Answer 1st step All …

Webf (x) = x^2+4kx+ (3+11k) where k is a constant 1. (I have no problem with this question but i'll just mention it) Express f (x) in the form (x+p)^2 + q, where p andq are constants …

Web15. The equation , where k is a constant, has 2 different real solutions for x. (a) Show that k satisfies (3) (b) Hence find the set of possible values of k. (4) (Total 7 marks) 2 16. f(x) = x + 4kx + (3 + 11k), where k is a constant. 2 (a) Express f(x) in the form (x + p) + q, where p and q are constants to be found in terms of k. (3) Given ... ezekiel jason hadiWebSep 16, 2024 · fx=x2+4kx+3+11k, where k is a constant.. a Express fx in the form x+p2+q , where p and q are constants to be found in terms of k. 3 Given that the equation fx=0 … hiasan batikWebf(x) = x2 + 4kx+ (3 + 11k); where k is a constant: (a) Express f(x) in the form (x+p)2 +q, where p and q are constants to be found in terms of k. [3] Given that the equation f(x) = 0 … hiasan bendera di pejabat