Leaving Cert Mathematics Algebra: Find the set of all real values

Find the set of all real values of $x$ for which $2x^2 + x - 15 \geq 0$ - Leaving Cert Mathematics - Question 2 - 2013

Question 2

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Find the set of all real values of $x$ for which $2x^2 + x - 15 \geq 0$. Solve the simultaneous equations; \[ x + y + z = 16 \] \[ \frac{5}{2}x + y + 10z = 40 \] \[... show full transcript

Worked Solution & Example Answer:Find the set of all real values of $x$ for which $2x^2 + x - 15 \geq 0$ - Leaving Cert Mathematics - Question 2 - 2013

Step 1

Find the set of all real values of $x$ for which $2x^2 + x - 15 \geq 0$.

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Answer

To solve the inequality $2x^2 + x - 15 \geq 0$ , we start by finding the roots of the equation:

$2x^2 + x - 15 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 2$ , $b = 1$ , and $c = -15$ ,

$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-15)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 120}}{4} = \frac{-1 \pm \sqrt{121}}{4} = \frac{-1 \pm 11}{4}$

Calculating the two possible values:

$x = \frac{10}{4} = \frac{5}{2}$
$x = \frac{-12}{4} = -3$

Now, we have the roots $x = -3$ and $x = \frac{5}{2}$ . These roots divide the number line into intervals: ((-\infty, -3), (-3, \frac{5}{2}), (\frac{5}{2}, \infty)$. We will test a point from each interval to determine where the quadratic is positive.

Test $x = -4$ :
$2(-4)^2 + (-4) - 15 = 32 - 4 - 15 = 13 \geq 0$ (True)
Test $x = 0$ :
$2(0)^2 + (0) - 15 = -15 < 0$ (False)
Test $x = 3$ :
$2(3)^2 + (3) - 15 = 18 + 3 - 15 = 6 \geq 0$ (True)

Thus, the solution set where $2x^2 + x - 15 \geq 0$ is:

$\{ x | x \leq -3 \} \cup \{ x | x \geq \frac{5}{2} \}$

Step 2

Solve the simultaneous equations; \[ x + y + z = 16 \] \[ \frac{5}{2}x + y + 10z = 40 \] \[ 2x + \frac{1}{2}y + 4z = 21. \]

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Answer

We start with the three equations:

( x + y + z = 16 )
( \frac{5}{2}x + y + 10z = 40 )
( 2x + \frac{1}{2}y + 4z = 21. )

From equation 1: $z = 16 - x - y$

Substituting $z$ into equation 2: $\frac{5}{2}x + y + 10(16 - x - y) = 40$ This simplifies to: $\frac{5}{2}x + y + 160 - 10x - 10y = 40$ $\frac{5}{2}x - 10x - 9y = -120$

Multiplying through by 2 to eliminate the fraction: $5x - 20x - 18y = -240$ $-15x - 18y = -240$ Thus, $5x + 6y = 80 \tag{Equation 4}$

Now, substitute $z$ into equation 3: $2x + \frac{1}{2}y + 4(16 - x - y) = 21$ This simplifies to: $2x + \frac{1}{2}y + 64 - 4x - 4y = 21$ $-2x - \frac{7}{2}y + 64 = 21$

Multiplying through by 2: $-4x - 7y = -86$ Thus, $4x + 7y = 86 \tag{Equation 5}$

Now we can solve equations 4 and 5 simultaneously: From equation 4: $x + \frac{6}{5}y = 16 \Rightarrow x = 16 - \frac{6}{5}y$ Now substitute into equation 5: $4(16 - \frac{6}{5}y) + 7y = 86$ Expanding: $64 - \frac{24}{5}y + 7y = 86$ Bringing all terms involving $y$ to one side: $64 + 7y - \frac{24}{5}y = 86$ $7y - \frac{24}{5}y = 86 - 64$ $7y - \frac{24}{5}y = 22$ Combine the $y$ terms: $\frac{35y - 24y}{5} = 22 \Rightarrow \frac{11y}{5} = 22$ $11y = 110 \Rightarrow y = 10$

Now substitute $y = 10$ back into equation 1 to find $x$ and $z$ : $x + 10 + z = 16 \Rightarrow z = 6 - x$ Substituting into equation 4: $5x + 6(10) = 80 \Rightarrow 5x + 60 = 80 \Rightarrow 5x = 20 \Rightarrow x = 4.$

Finally, substituting back to find $z$ : $z = 16 - 4 - 10 = 2.$ Thus, the solution set for the simultaneous equations is:

$x = 4, \; y = 10, \; z = 2.$

Find the set of all real values of $x$ for which $2x^2 + x - 15 \geq 0$ - Leaving Cert Mathematics - Question 2 - 2013

Worked Solution & Example Answer:Find the set of all real values of $x$ for which $2x^2 + x - 15 \geq 0$ - Leaving Cert Mathematics - Question 2 - 2013

Find the set of all real values of $x$ for which $2x^2 + x - 15 \geq 0$.

Solve the simultaneous equations; \[ x + y + z = 16 \] \[ \frac{5}{2}x + y + 10z = 40 \] \[ 2x + \frac{1}{2}y + 4z = 21. \]

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