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The quadratic equation x^2 - 2kx + (k - 1) = 0 has roots α and β such that α^2 + β^2 = 4. Without solving the equation, find the possible values of the real number k.

We know in a quadratic x^2 +bx + c = 0, -b/a = α + β and c/a = αβ. Therefore, α + β = -(-2k) = 2k, and αβ = k - 1. (Both are divided by the coefficient in front of x which is 1 so can be ignored. Now (α + β)...
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Answered by Ralph T. • Maths tutor
17718 Views

3. The point P lies on the curve with equation y=ln(x/3) The x-coordinate of P is 3. Find an equation of the normal to the curve at the point P in the form y = ax + b, where a and b are constants.

P- (3,0) y=ln(x/3) u=x/3 y=ln(u) ​​​​​​ du = 1/3 dy = 1/u = 3 dx du dy = du x dy dx dx du = 1/3 x 3 = 1 gradient at normal = -1 equation at normal = y = m(x) + c 0 = -3 + c 3 = c Answer: equation at normal =...
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Answered by Kaushalya B. • Maths tutor
13725 Views

In a sale a bag is reduced by 30%. The bag is now £31.50. Work out the original price of the bag.

£45.00
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Answered by Safiya C. • Maths tutor
3986 Views

Solve this simultaneous equation: (1) 2x+3y=12 (2) x+4y=11

x=3,y=2
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Answered by Safiya C. • Maths tutor
4171 Views

How do you solve simultaneous equations?

Simultaneous equations have two unknowns, you need to find out one of the unkowns (eg. x) in terms of the other (e.g. y) in order to solve both unknowns. Using these two examples (x and y) we can construct a...
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Answered by Charlie O. • Maths tutor
4057 Views