How do you solve $10 x^{2} + 9 = 499$ $10x^2+9=499$ ?

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How do you solve $10 x^{2} + 9 = 499$ $10x^2+9=499$ ?

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Answer 1 · 2018-03-29T00:08:38

$x = \pm 7$ $x = +- 7$

Explanation:

Given $10 x^{2} + 9 = 499$ $10x^2+9=499$ Solve for $x$ $x$

1) Subtract $9$ $9$ from both sides to isolate the $10 x^{2}$ $10x^2$ term

$10 x^{2} = 490$ $10x^2=490$

2) Divide both sides by $10$ $10$ to make the numbers smaller

$x^{2} = 49$ $x^2 = 49$

3) Find the square roots of both sides

$x = \pm 7$ $x = +- 7$ $\leftarrow$ $larr$ answer

---------- Check ----------

Sub in $7$ $7$ (or $- 7$ $-7$ ) in the place of $x$ $x$ in the original equation

$10 x^2 +9=499$
$10 (7^2)+9=499$ ?

Clear the exponent by squaring the $7$
$10(49) + 9 = 499$ ?

Clear the parentheses
$490 + 9 = 499$ ?

Combine like terms
$499 = 499$

$Check$

Answer 2 · 2018-03-29T00:11:22

x1 = 7.53
x2 = - 6.63

Explanation:

y = 10x^2 + 9x = 499
Use the improved quadratic formula (Socratic search):
$D = d^2 = b^2 - 4ac = 81 + 19,960 = 20,041$ .
$d = +- 141.57$
There are 2 real roots:
$x = -b/(2a) +- d/(2a) = - 9/20 +- 141.57/20$
$x = (9 +- 141.57)/20$
$x1 = 150.57/20 = 7.53$
$x2 = - 132.57/20 = - 6.63$
graph{10x^2 +9x - 499 [-320, 320, -160, 160]}

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How do you solve $10 x^{2} + 9 = 499$ $10x^2+9=499$ ?

2 Answers

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How do you solve $10 x^{2} + 9 = 499$ $10x^2+9=499$ ?