How do you solve $x^{2} - 9 x + 8 = 0$ $x^2-9x+8=0$ ?

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Answer 1 · 2016-03-12T06:33:48

$x = 8, 1$ $x=8,1$

$x^{2} - 9 x + 8 = 0$ $color(blue)(x^2-9x+8=0$

This is a Quadratic equation (in form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ )

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

Where

$a = 1, b = - 9, c = 8$ $color(blue)(a=1,b=-9,c=8$

Substitute the values in the formula

$\to x = \frac{- (- 9) \pm \sqrt{- 9^{2} - 4 (1) (8)}}{2 (1)}$ $rarrx=(-(-9)+-sqrt(-9^2-4(1)(8)))/(2(1))$

$\to x = \frac{9 \pm \sqrt{81 - 32}}{2}$ $rarrx=(9+-sqrt(81-32))/(2)$

$\to x = \frac{9 \pm \sqrt{49}}{2}$ $rarrx=(9+-sqrt(49))/(2)$

$\to x = \frac{9 \pm 7}{2}$ $rarrx=(9+-7)/(2)$

Now, we have $2$ $2$ values for $x$ $x$

$1) x = \frac{9 + 7}{2}$ $1)color(orange)(x=(9+7)/2$

$2) x = \frac{9 - 7}{2}$ $2)color(indigo)(x=(9-7)/2$

Solve

$1) x = \frac{9 + 7}{2} = \frac{16}{2} = 8$ $1)color(green)(x=(9+7)/2=16/2=8$

$2) x = \frac{9 - 7}{2}$ $2)color(green)(x=(9-7)/2$

How do you solve x^2-9x+8=0x2−9x+8=0 x^2-9x+8=0?