How do you solve $a^{2} + a - 30 = 0$ $a^2+a-30=0$ ?

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How do you solve $a^{2} + a - 30 = 0$ $a^2+a-30=0$ ?

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Answer 1 · 2015-07-08T05:47:07

Solve $y = x^{2} + x - 30 = 0$ $y = x^2 + x - 30 = 0$

Explanation:

$y = x^{2} + x - 30 = 0$ $y = x^2 + x - 30 = 0$
Find 2 numbers knowing sum (-1) and product (-30). Roots have opposite signs.
Factor pairs of (-30) --> ....(-3, 10)(-5, 6). This sum is 1 = b.
Then the 2 real roots are the opposite: $5 and - 6$ $5 and -6$

Answer 2 · 2015-07-08T06:38:13

Factor the trinomial into two binomials. Set each binomial equal to zero and solve for $a$ $a$ .

Explanation:

$a^{2} + a - 30 = 0$ $a^2+a-30=0$

Find two numbers that when added equal $1$ $1$ , and when multiplied equal $- 30$ $-30$ .

The numbers $- 5$ $-5$ and $6$ $6$ fit the pattern.

$a^{2} + a - 30 = 0$ $a^2+a-30=0$ =

$(a - 5) (a + 6) = 0$ $(a-5)(a+6)=0$

Set $(a - 5)$ $(a-5)$ equal to zero and solve for $a$ $a$ .

$a - 5 = 0$ $a-5=0$

$a = 5$ $a=5$

Set $(a + 6)$ $(a+6)$ equal to zero and solve for $a$ $a$ .

$a + 6 = 0$ $a+6=0$

$a = - 6$ $a=-6$

$a = - 6, 5$ $a=-6, 5$

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How do you solve $a^{2} + a - 30 = 0$ $a^2+a-30=0$ ?

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How do you solve a2+a−30=0a^2+a-30=0?

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How do you solve $a^{2} + a - 30 = 0$ $a^2+a-30=0$ ?