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# 1. factorize 9(2a-b)^2 - 4(2a-b)^2 -13

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1.  factorize 9(2a-b)^2 - 4(2a-b)^2 -13

2. if p = 2-a, then find the value of a^3 + 6ap + p^3 -8

3. if the parallel sides of trapezium are 25cm and 11cm, while its non- parallel sides are 15cm and 13 cm . find the area of the trapezium...

kes1968  Jul 12, 2015

#5
+20680
+5

2. if p = 2-a, then find the value of a^3 + 6ap + p^3 -8

$$\small{\text{ \begin{array}{rclcl} a^3 + 6ap + p^3 -8 \\ &=& a^3 + p^3 + 6ap -8 \qquad &|& \qquad (a^3+p^3) = (a+p)(a^2-ap+p^2)\\ &=& (a+p)(a^2-ap+p^2)+ 6ap -8 \qquad &|& \qquad p = 2-a \\ &=& (a+2-a)(a^2-ap+p^2)+ 6ap -8 \\ &=& 2(a^2-ap+p^2)+ 6ap -8 \\ &=& 2\left[ a^2-ap+p^2+ 3ap -4\right] \\ &=& 2( a^2+2ap+p^2 -4) \qquad &|& \qquad (a^2+2ap+p^2) = (a+p)^2 \\ &=& 2\left[ (a+p)^2 -4 \right] \qquad &|& \qquad p = 2-a \\ &=& 2\left[ (a+2-a)^2 -4) \right]\\ &=& 2( 2^2 -4)\\ &=& 2( 4 -4)\\ &=& 2\cdot 0\\ &=& 0 \end{array} }}$$

heureka  Jul 13, 2015
#1
+94120
+5

Hi Kes,

$$\\9(2a-b)^2 - 4(2a-b)^2 -13\\ 5(2a-b)^2-13\\ 5(4a^2-4ab+b^2)-13\\ 20a^2-20ab+5b^2-13\\$$

Mmm     I don't think that this can be factorized...

I suppose that it is the difference of 2 squares so it is equal to

$$(\sqrt5(2a-b)-\sqrt{13})(\sqrt5(2a-b)+\sqrt{13})$$

but I don't think this looks very nice....

Melody  Jul 12, 2015
#2
+94120
+5

2. if p = 2-a, then find the value of a^3 + 6ap + p^3 -8

$$\\a^3 + 6ap + p^3 -8 \\ = (6ap+p^3)-(8-a^3)\\ =p(6a+p^2)-(2-a)(4+2a+a^2)\\ sub\;\; 2-a for p\\ =p(6a+p^2)-p(4+2a+a^2)\\  factor out the p\\ =p(6a+p^2-4-2a-a^2)\\ =p(p^2-a^2+4a-4)\\ =p((2-a)^2-a^2+4a-4)\\ =p(4+a^2-4a-a^2+4a-4)\\ =p(0)\\ =0$$

Melody  Jul 12, 2015
#3
+94120
+5

3. if the parallel sides of trapezium are 25cm and 11cm, while its non- parallel sides are 15cm and 13 cm . find the area of the trapezium...

I am let the long parallel side 25cm be the bottom.

15+13=28  Only 3cm more than 25cm.

So it is obvious to me that the two angles at the bottom MUST both be acute angles.

Here is the trapezium.

Now I am going to cut the two triangles of the end and push them up together so that I have a triangle with side lengths   14cm on the bottom and 15cm and 13cm on the other two sides

The height of the triangle is also the height of the trapezium.

Here is a pic

Using pythagoras's theorem

$$\\15^2=(14-x)^2+h^2\\ 225=196+x^2-28x+h^2\\ 29=x^2-28x+h^2\qquad(1)\\\\ 169=x^2+h^2\qquad (2)\\ (2)-(1)\\ 140=28x\\ x=5\\ so\\ 169=25+h^2\\ 144=h^2\\ h=12cm\\\\ Area of trapezium  = \frac{(25+11)}{2}\times 12=216cm^2$$

Melody  Jul 12, 2015
#4
+20680
+5

1.  factorize 9(2a-b)^2 - 4(2a-b)^2 -13

$$\small{\text{ \begin{array}{rcl} 9x^2 - 4x^2 -13 &=& 5x^2 - 13 \qquad | \qquad x= 2a-b \\\\ 5x^2 - 13 &=& 0\\ 5x^2 &=& 13 \\\\ x^2 &=& \dfrac{13}{5}\\\\ x^2 &=& 2.6 \\ \mathbf{x_{1,2} } & \mathbf{=} & \mathbf{\pm\sqrt{2.6}}\\\\ 5x^2 - 13 &=& 5\cdot(x-\sqrt{2.6})(x+\sqrt{2.6})\qquad | \qquad x= 2a-b \\\\ 9(2a-b)^2 - 4(2a-b)^2 -13 = 5(2a-b)^2 - 13 &=& 5\cdot\left[(2a-b)-\sqrt{2.6}\right] \left[(2a-b)+\sqrt{2.6} \right] \\\\ \end{array} }}$$

heureka  Jul 13, 2015
#5
+20680
+5

2. if p = 2-a, then find the value of a^3 + 6ap + p^3 -8

$$\small{\text{ \begin{array}{rclcl} a^3 + 6ap + p^3 -8 \\ &=& a^3 + p^3 + 6ap -8 \qquad &|& \qquad (a^3+p^3) = (a+p)(a^2-ap+p^2)\\ &=& (a+p)(a^2-ap+p^2)+ 6ap -8 \qquad &|& \qquad p = 2-a \\ &=& (a+2-a)(a^2-ap+p^2)+ 6ap -8 \\ &=& 2(a^2-ap+p^2)+ 6ap -8 \\ &=& 2\left[ a^2-ap+p^2+ 3ap -4\right] \\ &=& 2( a^2+2ap+p^2 -4) \qquad &|& \qquad (a^2+2ap+p^2) = (a+p)^2 \\ &=& 2\left[ (a+p)^2 -4 \right] \qquad &|& \qquad p = 2-a \\ &=& 2\left[ (a+2-a)^2 -4) \right]\\ &=& 2( 2^2 -4)\\ &=& 2( 4 -4)\\ &=& 2\cdot 0\\ &=& 0 \end{array} }}$$

heureka  Jul 13, 2015