How do you solve $y^{2} - 15 y + 54 \geq 0$ $y^{ 2} - 15y + 54\geq 0$ ?

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How do you solve $y^{2} - 15 y + 54 \geq 0$ $y^{ 2} - 15y + 54\geq 0$ ?

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Answer 1 · 2017-12-02T08:59:02

Solutions:
$y \leq 6 or y \geq 9$ $y <= 6 or y >=9$

Using Interval Notations :
$(- \infty, 6] \cup [9, \infty)$ $(-oo, 6] uu [9, oo)$

enter image source here

Explanation:

Graph attached as visual proof of our required solutions.

We have the following Quadratic Inequality given to us:

$y^{2} - 15 y + 54 \geq 0$ $y^2 - 15y + 54 >= 0$

We can factorize the quadratic expression on the left-hand side as follows and rewrite our quadratic inequality :

$y^{2} - 9 y - 6 y + 54 \geq 0$ $y^2 - 9y - 6y + 54 >= 0$

$\Rightarrow y (y - 9) - 6 (y - 9) \geq 0$ $rArr y(y -9) -6(y - 9) >= 0$

Therefore, $(y - 9) (y - 6) \geq 0$ $(y -9)(y - 6) >= 0$

Hence, $(y - 9) = 0 or (y - 6) = 0$ $(y - 9) =0 or (y -6 ) =0$

We get two values for $y$ $y$ .

They are $y = 9, y = 6$ $y =9, y = 6$

Next step is to choose values for $y$ $y$ as follows, to test them on the Number Line:

A value less than 6; a value $> 6$ $>6$ but $< 9$ $<9$ and a value $> 9$ $>9$ to verify whether our inequality $y^{2} - 15 y + 54 \geq 0$ $y^2 - 15y + 54 >= 0$ works for these values.

Let these values by $y = 5$ $y =5$ ; $y = 7$ $" "y =7$ and $y = 10$ $y = 10$

When $y = 5$ $color(red)(y = 5)$

$5^{2} - 15 (5) + 54 \geq 0$ $5^2 - 15(5) + 54 >= 0$

$25 - 75 + 54 \geq 0$ $25 - 75 + 54 >=0$

$\Rightarrow 4 \geq 0$ $rArr4 >=0$

We observe that the value $y = 5$ $" "y=5$ satisfies our inequality

When $y = 7$ $color(red)(y = 7)$

$7^{2} - 15 (7) + 54 \geq 0$ $7^2 - 15(7) + 54 >= 0$

$49 - 105 + 54 \geq 0$ $49 - 105 + 54 >=0$

$\Rightarrow - 2 \geq 0$ $rArr-2 >=0$

We observe that the value $y = 7$ $" "y=7$ does not satisfy our inequality. Hence we must keep this fact in our mind when we indicate our intervals for the inequality.

When $y = 10$ $color(red)(y = 10)$

$10^{2} - 15 (10) + 54 \geq 0$ $10^2 - 15(10) + 54 >= 0$

$100 - 150 + 54 \geq 0$ $100 - 150 + 54 >=0$

$\Rightarrow 4 \geq 0$ $rArr4 >=0$

We observe that the value $y = 10$ $" "y=10$ satisfies our inequality.

Hence, our solutions to the inequality $y^{2} - 15 y + 54 \geq 0$ $y^2 - 15y + 54 >= 0$ are given by:

$y \leq 6 or y \geq 9$ $color(blue)(y <= 6 or y >=9)$

Using Interval Notations our solution set is :

$(- \infty, 6] \cup [9, \infty)$ $color(green)((-oo, 6] uu [9, oo))$

Next, we use a Number Line and mark these values of $y$ $y$

Please refer to the Image attached for the Number Line.

The graph below will provide a visual evidence of our findings:

enter image source here

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How do you solve $y^{2} - 15 y + 54 \geq 0$ $y^{ 2} - 15y + 54\geq 0$ ?

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How do you solve y2−15y+54≥0y^{ 2} - 15y + 54\geq 0?

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Explanation:

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How do you solve $y^{2} - 15 y + 54 \geq 0$ $y^{ 2} - 15y + 54\geq 0$ ?