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(a) (i) Factorise $x^2 + 7x - 30$ - Junior Cycle Mathematics - Question 9 - 2015

Question 9

(a) (i) Factorise $x^2 + 7x - 30$. (ii) Hence, or otherwise, solve the equation $x^2 + 7x - 30 = 0$. (b) Solve the equation $2x^2 - 7x - 10 = 0$. Give each answer ... show full transcript

Worked Solution & Example Answer:(a) (i) Factorise $x^2 + 7x - 30$ - Junior Cycle Mathematics - Question 9 - 2015

Step 1

Factorise $x^2 + 7x - 30$

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Answer

To factorise the quadratic expression $x^2 + 7x - 30$ , we need to find two numbers that multiply to -30 and add to +7. The numbers +10 and -3 satisfy these conditions. Thus, we can express the quadratic as:

$x^2 + 7x - 30 = (x + 10)(x - 3)$

Step 2

Hence, or otherwise, solve the equation $x^2 + 7x - 30 = 0$

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Answer

Using the factored form $(x + 10)(x - 3) = 0$ , we can set each factor to zero:

$x + 10 = 0 ightarrow x = -10$
$x - 3 = 0 ightarrow x = 3$

Thus, the solutions are $x = 10$ or $x = -3$ .

Step 3

Solve the equation $2x^2 - 7x - 10 = 0$

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Answer

To solve the equation $2x^2 - 7x - 10 = 0$ , we can either factor or use the quadratic formula. Here, we will use the quadratic formula:

$x = rac{-b m{ ext{±}} ext{sqrt}(b^2 - 4ac)}{2a}$

Substituting $a = 2$ , $b = -7$ , and $c = -10$ ,

$x = rac{7 m{ ext{±}} ext{sqrt}((-7)^2 - 4(2)(-10))}{2(2)}$ $= rac{7 m{ ext{\pm}} ext{sqrt}(49 + 80)}{4}$ $= rac{7 m{ ext{\pm}} ext{sqrt}(129)}{4}$

Calculating the two potential answers:

$x ext{ (positive)} = rac{7 + ext{sqrt}(129)}{4} \approx 4.59$
$x ext{ (negative)} = rac{7 - ext{sqrt}(129)}{4} \approx -1.09$

Thus, the solutions to two decimal places are $4.59$ and $-1.09$ .

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(a) (i) Factorise $x^2 + 7x - 30$ - Junior Cycle Mathematics - Question 9 - 2015

Worked Solution & Example Answer:(a) (i) Factorise $x^2 + 7x - 30$ - Junior Cycle Mathematics - Question 9 - 2015

Factorise $x^2 + 7x - 30$

Hence, or otherwise, solve the equation $x^2 + 7x - 30 = 0$

Solve the equation $2x^2 - 7x - 10 = 0$

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