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Please expand the brackets in the following equation to get a quadratic equation. Then, please show using the quadratic formula that the solutions to the equation are x=3 and x=5. Here is the starting equation: (x-3)(x-5)=0

(X-3)(x-5)=0Use F.O.I.LFirst x multiplied by x gives x²outer x multiplied by -5 gives -5xinner x multiplied by -3 gives -3xlast -5x-3 gives +15combining we get x²-8x+15=0the quadrati...

Answered by Jacob G. • Maths tutor

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Solve 7x + 6 > 1 + 2x

7x + 6 > 1 + 2xFirst, we collect the like terms so we move all the x's to one side and all the integers to the other:5x > -5Then we divide by 5 on each side to find what just x will be.Therefore, x ...

Answered by Juliet Z. • Maths tutor

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Solve the simultaneous equation, 3x + y = 8 and x + 3y = 12, to find a value for x and y.

Re arrange one of the equation to get a single variant answer. So, x = 12 - 3y. Substitute this into the other equation, so 3 (12-3y) + y = 8. Expand this equation to form 36 - 9y + y = 8. Collect the y t...

Answered by Grace R. • Maths tutor

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Show that (x + 4)(x + 5)(x + 6) can be written in the form ax3 + bx2 + cx + d where a, b, c and d are positive integers.

=(x+4)(x²+6x+5x+30)=(x+4)(x²+11x+30)=x³+11x²+30x+4x²+44x+120=x³+15x²+74x+120

Answered by Jiawen Z. • Maths tutor

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The Diagram shows the Triangle PQR. PQ = x cm. PR = 2x cm. Angle QP^R = 30 degrees. The area of the triangle PQR = A cm^2. Show that x = (Squared Root){2A

Area of a Triangle Formula is
A = 1/2 abSINc
Label the sides of the triangle PR = 2x = a PQ = x = b
1/2 (x) 2x (x) x (x) SINc = A = x²SINc
Rearrange the equation to give...

Answered by Tutor305386 D. • Maths tutor

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