Sophie D.

Physics Student at King's College London

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Pre-Calculus

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Question:

Find the roots of the equation y = x^2 - 2x - 3. Explain what this would look like on a graph.

Sophie D.

Answer:

To find the roots of this equation, we need to think what two things we would multiply together to give us the equation. Because it is a quadratic, we know it would be two factors in the form (x+a) and (x+b), where a and b are numbers. In this case, that turns out to be (x+1) and (x-3). Try multiplying these two together yourself to convince yourself this is correct. On a graph, firstly the graph would be a u shape; that is, a quadratic curve. We also know that this curve would cross the x axis when x = -1, and x =3. This is because those values, when plugged into our root terms, would make 0, so y=0, so we are on the x axis. If you wanted to know what value the curve crosses the y axis, simply make x=0 in our original equation; so in this case, it would cross the y axis at y=-3.

Pre-Algebra

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Question:

Simplify the following equation in terms of x and y: 5x + 5y + 15x - 4y And state what the terms and coefficients of the final equations are.

Sophie D.

Answer:

First we will look at grouping together the parts of the equation that contain x. Do this by looking at which numbers have x next to them and adding the numbers together. This gives us: 20x + 5y - 4y Then group together the parts that contain y. Handle this the same way you handled the parts contianing x. This gives us: 20x - y The terms in this equation are the letters; x and y. The coefficients are the numbers in front of the terms. So, 20 would be the coefficient of x, and -1 would be the coefficient of y.

Algebra

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Question:

Solve the equation for x: [10(5x+2)]/[3x^2+3x]=(1/5)(5+20x^-1)

Sophie D.

Answer:

First, expand out and simplify all brackets. This gives us: [50x+20]/[3x^2+3x]=(1+4) Then, multiply both sides by the denominator of the left hand side ([3x^2+3x]), giving us: 50x+20=(3x^2+3x)(5) simplify again and put into the form of a quadratic equation (so one of the form ax^2+bx+c=0, where a,b and c are all integer values). We therefore get: 15x^2-35x-20=0 We then divide through by 15 so that x^2 has coefficient of value 1. Then use the complete the square method to arrive at two possible solutions of x, by the following steps: 15x^2-35x-20= 0 goes to x^2-(35/15)x-(20/15)=0 Then in complete the square form = (x-(35/30))^2 - (49/36)-(20/15)=0 simplify to get = (x-(7/6))^2 - (97/36) =0 then: x - (7/6) = +/- [(square root of (97))/6] so our values of x are 2.81 and -15.

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