Hugh K.

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Mmath - G103 - Mathematics (Masters) - Warwick University

4.7

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4.7from 4 customer reviews

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October 3 2016

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#### Qualifications

MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A
ChemistryA-level (A2)A
PhysicsA-level (A2)A*

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#### Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£20 /hr
MathsA Level£20 /hr
ChemistryGCSE£18 /hr
Further MathematicsGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr
Further MathematicsIB£20 /hr
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### For what values of k does the graph y=x^(2)+2kx+5 not intersect the x-axis

Where the graph intersects the x-axis, x2+2kx+5 must be equal to zero. Thus we can answer the equivalent question: For what k does x2+2kx+5 = 0 not have a solution?

This is now a simpler problem (roots of a quadratic equation). We can apply the common method of considering the discriminant of x2+2kx+5. Using standard quadratic formula notation where in this case a=1, b =2k and c=5 we evaluate the discriminant : b2-4ac= (2k)2-4*1*5 = 4k2 -20.

Now since the discriminant appears in a square root sign in the quadratic equation, if it is negative there can be no real solutions to the equation ( great this is what we want!).

Thus we want discriminant negative: 4k2 -20 <0. Divide both sides of the inequality by 4 so we have k2-5<0.

Now this is where we must take great care, the following reasoning is a common MISTAKE: rearragne the inequality so we have k2 < 5, then squarrot both sides so we have k < sqrt(5) or k < - sqrt(5) . The second inequalit is implied by the first thus the discriminate negtive for all k values les then the sqrt(5). THIS IS INCORRECT.

When dealing with inequalities involving powers such as we are here we must be extremely careful. the mistake in the reasoning above is when we say k < - sqrt(5), this is actually a form of the common mistake of not inverting the inequality when multiplying both sides of an equation by a negative. Instead when dealing with inequalities with powers it is always much wiser to sketch a graph of the situation.

k2 - 5 is the standard quadratic U shape (think y=x2) shifted down by 5. Having sketched this out it is clear that this graph is less then 0 when it is inbetween it's two roots.

The roots of k-5 are easy to find: k-5 = 0 implies k2 = 5 implies k = sqrt(5) or k = -sqrt(5).

Comparing this with the graph we can now see that the discriminant is negative for - sqrt(5) < k <  sqrt(5). Thus these are the values for which the graph y=x2+2kx+5 does not intersect the x-axis.

Where the graph intersects the x-axis, x2+2kx+5 must be equal to zero. Thus we can answer the equivalent question: For what k does x2+2kx+5 = 0 not have a solution?

This is now a simpler problem (roots of a quadratic equation). We can apply the common method of considering the discriminant of x2+2kx+5. Using standard quadratic formula notation where in this case a=1, b =2k and c=5 we evaluate the discriminant : b2-4ac= (2k)2-4*1*5 = 4k2 -20.

Now since the discriminant appears in a square root sign in the quadratic equation, if it is negative there can be no real solutions to the equation ( great this is what we want!).

Thus we want discriminant negative: 4k2 -20 <0. Divide both sides of the inequality by 4 so we have k2-5<0.

Now this is where we must take great care, the following reasoning is a common MISTAKE: rearragne the inequality so we have k2 < 5, then squarrot both sides so we have k < sqrt(5) or k < - sqrt(5) . The second inequalit is implied by the first thus the discriminate negtive for all k values les then the sqrt(5). THIS IS INCORRECT.

When dealing with inequalities involving powers such as we are here we must be extremely careful. the mistake in the reasoning above is when we say k < - sqrt(5), this is actually a form of the common mistake of not inverting the inequality when multiplying both sides of an equation by a negative. Instead when dealing with inequalities with powers it is always much wiser to sketch a graph of the situation.

k2 - 5 is the standard quadratic U shape (think y=x2) shifted down by 5. Having sketched this out it is clear that this graph is less then 0 when it is inbetween it's two roots.

The roots of k-5 are easy to find: k-5 = 0 implies k2 = 5 implies k = sqrt(5) or k = -sqrt(5).

Comparing this with the graph we can now see that the discriminant is negative for - sqrt(5) < k <  sqrt(5). Thus these are the values for which the graph y=x2+2kx+5 does not intersect the x-axis.

2 years ago

4092 views

### If cos(x)= 1/3 and x is acute, then find tan(x).

Consider a right angled triangle. Call one of the angles (not the right angle) in this triangle x. We can do this as we are told x is acute. The side opposite to x label O, the side adjacent to x label A, and label the hypotenuse H.

Now from SOHCAHTOA cos(x) = A/H = 1/3 and tan(x) = O/A . We also know by pythagoras that A+ O2 = H2 . We shall now combine these equations to get our result.

A/H = 1/3 implies H = 3A implies H2 = 9A2. Substituting this result into our euation obtained by pythagoras we get: A+ O2 = 9A2. Rearranging: O2 = 8A2 implies O2/A2 = 8 implies (O/A)2 = 8. Now we take the square root of both sides. Here we must take care, O and A are lengths and so are not negative, so we only consider the positive root: O/A = sqrt(8) = tan(x) and so we are done.

Consider a right angled triangle. Call one of the angles (not the right angle) in this triangle x. We can do this as we are told x is acute. The side opposite to x label O, the side adjacent to x label A, and label the hypotenuse H.

Now from SOHCAHTOA cos(x) = A/H = 1/3 and tan(x) = O/A . We also know by pythagoras that A+ O2 = H2 . We shall now combine these equations to get our result.

A/H = 1/3 implies H = 3A implies H2 = 9A2. Substituting this result into our euation obtained by pythagoras we get: A+ O2 = 9A2. Rearranging: O2 = 8A2 implies O2/A2 = 8 implies (O/A)2 = 8. Now we take the square root of both sides. Here we must take care, O and A are lengths and so are not negative, so we only consider the positive root: O/A = sqrt(8) = tan(x) and so we are done.

2 years ago

1614 views

### If n is an integer prove (n+3)^(2)-n^(2) is never even.

Let us begin by simplifying the expression:

(n+3)2 - n2  = (n+3)(n+3) - n2

= n2 + 6n + 9 - n2 (expanded brackets)

= 6n + 9 (collected like terms)

= 3(2n+3) (taken out a factor of 3)

Now we can consider this simpler equivalent expression.

3 is an odd number

2n is even thus 2n+3 is odd (even plus odd is always odd)

so we have an odd*odd which is always odd, thus never even and we are done.

Let us begin by simplifying the expression:

(n+3)2 - n2  = (n+3)(n+3) - n2

= n2 + 6n + 9 - n2 (expanded brackets)

= 6n + 9 (collected like terms)

= 3(2n+3) (taken out a factor of 3)

Now we can consider this simpler equivalent expression.

3 is an odd number

2n is even thus 2n+3 is odd (even plus odd is always odd)

so we have an odd*odd which is always odd, thus never even and we are done.

2 years ago

822 views

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