# Tutor profile: Sophie D.

## Questions

### Subject: Pre-Calculus

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

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.

### Subject: Pre-Algebra

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.

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.

### Subject: Algebra

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

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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