How do you solve $x^{4} - 10 x^{2} + 9 \geq 0$ $x^4-10x^2+9>=0$ using a sign chart?

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How do you solve $x^{4} - 10 x^{2} + 9 \geq 0$ $x^4-10x^2+9>=0$ using a sign chart?

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Answer 1 · 2016-10-27T14:10:55

Solution is $- \infty \leq x \leq - 3$ $-oo <= x <= -3$ or $- 1 \leq x \leq 1$ $-1 <= x <= 1$ , or $3 \leq x \leq \infty$ $3 <= x <= oo$ and in interval form it can be written as $[- \infty, - 3] \cup [- 1, 1] \cup [3, \infty]$ $[-oo,-3]uu[-1,1]uu[3,oo]$

Explanation:

Let us first factorize $x^{4} - 10 x^{2} + 9$ $x^4-10x^2+9$ .

$x^{4} - 10 x^{2} + 9 = x^{4} - 9 x^{2} - x^{2} + 9$ $x^4-10x^2+9=x^4-9x^2-x^2+9$

= $x^{2} (x^{2} - 9) - 1 (x^{2} - 9) = (x^{2} - 9) (x^{2} - 1))$ $x^2(x^2-9)-1(x^2-9)= (x^2-9)(x^2-1))$

= $(x + 3) (x - 3) (x + 1) (x - 1)$ $(x+3)(x-3)(x+1)(x-1)$

Hence we have to solve the inequality

$(x + 3) (x - 3) (x + 1) (x - 1) \geq 0$ $(x+3)(x-3)(x+1)(x-1)>=0$

From this we know that the product $(x + 3) (x - 3) (x + 1) (x - 1)$ $(x+3)(x-3)(x+1)(x-1)$ has to be zero or positive. It is apparent that sign of binomials $(x + 3)$ $(x+3)$ , $(x + 1)$ $(x+1)$ , $(x - 1)$ $(x-1)$ and $(x - 3)$ $(x-3)$ will change around the values $- 3$ $-3$ . $- 1$ $-1$ , $1$ $1$ and $3$ $3$ respectively. In sign chart we divide the real number line using these values, i.e. below $- 3$ $-3$ , between $- 3$ $-3$ and $- 1$ $-1$ , between $- 1$ $-1$ and $1$ $1$ , between $1$ $1$ and $3$ $3$ and above $3$ $3$ and see how the sign of $(x + 3) (x - 3) (x + 1) (x - 1)$ $(x+3)(x-3)(x+1)(x-1)$ changes.

Sign Chart

$X X X X X X X X - 3 X X X X - 1 X X X X 1 X X X X 3$ $color(white)(XXXXXXXX)-3color(white)(XXXX)-1color(white)(XXXX)1color(white)(XXXX)3$

$(x + 3) X X - i v e X X + i v e X X X + i v e X X + i v e X X + i v e$ $(x+3)color(white)(XX)-ive color(white)(XX)+ive color(white)(XXX)+ive color(white)(XX)+ive color(white)(XX)+ive$

$(x + 1) X X - i v e X X - i v e X X X + i v e X X + i v e X X + i v e$ $(x+1)color(white)(XX)-ive color(white)(XX)-ive color(white)(XXX)+ive color(white)(XX)+ive color(white)(XX)+ive$

$(x - 1) X X - i v e X X - i v e X X X - i v e X X + i v e X X + i v e$ $(x-1)color(white)(XX)-ive color(white)(XX)-ive color(white)(XXX)-ive color(white)(XX)+ive color(white)(XX)+ive$

$(x - 3) X X - i v e X X - i v e X X X - i v e X X - i v e X X + i v e$ $(x-3)color(white)(XX)-ive color(white)(XX)-ive color(white)(XXX)-ive color(white)(XX)-ive color(white)(XX)+ive$

$x^{4} - 10 x^{2} + 9$ $x^4-10x^2+9$
$X X X X X X + i v e X x - i v e X X X + i v e X X - i v e X X + i v e$ $color(white)(XXXXXX)+ive color(white)(Xx)-ive color(white)(XXX)+ive color(white)(XX)-ive color(white)(XX)+ive$

It is observed that $x^{4} - 10 x^{2} + 9 \geq 0$ $x^4-10x^2+9 >= 0$

when either $- \infty \leq x \leq - 3$ $-oo <= x <= -3$ or $- 1 \leq x \leq 1$ $-1 <= x <= 1$ , or $3 \leq x \leq \infty$ $3 <= x <= oo$ , which is the solution for the inequality.

In interval notation, this can be written as $[- \infty, - 3] \cup [- 1, 1] \cup [3, \infty]$ $[-oo,-3]uu[-1,1]uu[3,oo]$

Answer 2 · 2016-10-27T14:53:43

The values of x are
$- \infty \leq x \leq - 3$ $-oo<=x<=-3$ and $- 1 \leq x \leq 1$ $-1<=x<=1$

Explanation:

Let's start by factorising the expression

$x^{4} - 10 x^{2} + 9 = (x^{2} - 1) (x^{2} - 9) = (x + 1) (x - 1) (x + 3) (x - 3)$ $x^4-10x^2+9=(x^2-1)(x^2-9)=(x+1)(x-1)(x+3)(x-3)$

So we can make the sign chart
$x$ $x$ $a a a a a$ $color(white)(aaaaa)$ -oo $a a a a$ $color(white)(aaaa)$ -3 $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$
$x + 3$ $x+3$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ 0 $a a$ $color(white)(aa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a$ $color(white)(aa)$ $+$ $+$ $a a$ $color(white)(aa)$ $+$ $+$
$x + 1$ $x+1$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a$ $color(white)(a)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a$ $color(white)(aa)$ $+$ $+$
You can continue the sign chart
and you wiil find that
$x^{4} - 10 x^{2} + 9 \geq 0$ $x^4-10x^2+9>=0$ for $- \infty \leq x \leq - 3$ $-oo<=x<=-3$ and $- 1 \leq x \leq 1$ $-1<=x<=1$

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How do you solve $x^{4} - 10 x^{2} + 9 \geq 0$ $x^4-10x^2+9>=0$ using a sign chart?

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How do you solve $x^{4} - 10 x^{2} + 9 \geq 0$ $x^4-10x^2+9>=0$ using a sign chart?