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

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

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Answer 1 · 2016-04-11T03:19:45

1 and 9

Explanation:

$y = x^{2} - 10 x + 9 = 0 .$ $y = x^2 - 10x + 9 = 0.$
Find 2 real roots, knowing sum (-b = 10) and product (c = 9).
They are 1 and 9.

Answer 2 · 2016-04-11T09:32:28

$9, 1$ $9,1$

Explanation:

$x^{2} + 9 = 10 x$ $color(blue)(x^2+9=10x$

Subtract $10 x$ $10x$ both sides

$\to x^{2} + 9 - 10 x = 10 x - 10 x$ $rarrx^2+9-10x=cancel(10x-10x$

$\to x^{2} + 9 - 10 x = 0$ $rarrx^2+9-10x=0$

Rewrite the equation in Standard form ( $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ )

$\to x^{2} - 10 x + 9 = 0$ $color(purple)(rarrx^2-10x+9=0$

This is a Quadratic formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)$

Remember that $a and b$ $aand b$ are the coefficients and $c$ $c$ is the constant

So,

$a = 1$ $color(violet)(a=1$

$b = - 10$ $color(violet)(b=-10$

$c = 9$ $color(violet)(c=9$

$\to x = \frac{- (- 10) \pm \sqrt{- 10^{2} - 4 (1) (9)}}{2 (1)}$ $rarrx=(-(-10)+-sqrt(-10^2-4(1)(9)))/(2(1))$

$\to x = \frac{10 \pm \sqrt{100 - (36)}}{2}$ $rarrx=(10+-sqrt(100-(36)))/(2)$

$\to x = \frac{10 \pm \sqrt{64}}{2}$ $rarrx=(10+-sqrt(64))/(2)$

$\to x = \frac{10 \pm 8}{2}$ $rarrx=(10+-8)/(2)$

$\to x = \frac{10^{5} \pm 8^{4}}{2^{1}}$ $rarrx=(cancel10^5+-cancel8^4)/(cancel2^1)$

$\to x = 5 \pm 4$ $rarrx=5+-4$

Remember that $\pm$ $+-$ means "plus or minus",

Which implies that,

$x = 5 + 4 = 9$ $color(indigo)(x=5+4=9$

$x = 5 - 4 = 1$ $color(orange)(x=5-4=1$

$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ | x = 9, 1 | - ------ -$ $color(blue)( ul bar |x=9,1|$

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

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