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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)...
RT
Answered by Ralph T. • Maths tutor
17721 Views

Consider the arithmetic sequence 2, 5, 8, 11, ... a) Find U101 b) Find the value of n so that Un = 152

Firstly, as with any question, make sure to check your formula book in order to find any relevant equations. In this case, the one most relevant to us is U n = U 1 + d(n-1). From here we will need to find th...
KW
Answered by Kirsty W. • Maths tutor
12697 Views

IB exam question: Let p(x)=2x^5+x^4–26x^3–13x^2+72x+36, x∈R. For the polynomial equation p (x) = 0 , state (i) the sum of the roots; (ii) the product of the roots.

p(x) = 2x 5 + x 4 – 26x 3 – 13x 2 + 72x + 36 i) obtain the sum by using vieta's fomula: for p(x) the sum is hence -b/a hence, sum = -1/2 ii) obtain the product by using the fomula -f/a. This is because the p...
NM
Answered by Niccolo M. • Maths tutor
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What is the equation of the tangent drawn to the curve y = x^3 - 2x + 1 at x = 2?

The equation of a two-dimensional non-vertical line can easily be determined if its gradient 'm' and a point on the line (x0, y0) is known, using the formula m = (y - y0) / (x - x0). The gradient of the tang...
MZ
Answered by Mario Z. • Maths tutor
2242 Views

a) Let u=(2,3,-1) and w=(3,-1,p). Given that u is perpendicular to w, find the value of p. b)Let v=(1,q,5). Given that modulus v = sqrt(42), find the possible values of q.

a) u is perpendicular to w , so u • w =0. This means that 2 3 + 3 (-1) + (-1)*p = 0, so 6-3-p=0, hence p=3. b) The magnitude of v = sqrt(1 2 +q 2 +5 2 ) = sqrt(42). Thus, 1+q 2 +25 = 42, q 2 =16, so q is eit...
CR
Answered by Cristiana R. • Maths tutor
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