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3. A high school student is trying to solve the equation 2x^2+30=-9x They decides to use the quadratic formula to solve and their work is shown below. 2x^2+30=-9x a=2 b=30 x=(-30pm sqrt ((30)^2-4(2)(-9)))/(2(2)) x=(-30pm sqrt (900+72))/(4) x=(-30pm 18sqrt (3))/(4) x=(15)/(2)pm (9sqrt (3))/(2) Is the student correct?If so why, if not why? Explain your answer:

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3. A high school student is trying to solve the equation 2x^2+30=-9x They decides to use the quadratic formula to solve and their work is shown below. 2x^2+30=-9x a=2 b=30 x=(-30pm sqrt ((30)^2-4(2)(-9)))/(2(2)) x=(-30pm sqrt (900+72))/(4) x=(-30pm 18sqrt (3))/(4) x=(15)/(2)pm (9sqrt (3))/(2) Is the student correct?If so why, if not why? Explain your answer:

3. A high school student is trying to solve the equation 2x^2+30=-9x They decides to use the quadratic formula to solve and their work is shown below. 2x^2+30=-9x a=2 b=30 x=(-30pm sqrt ((30)^2-4(2)(-9)))/(2(2)) x=(-30pm sqrt (900+72))/(4) x=(-30pm 18sqrt (3))/(4) x=(15)/(2)pm (9sqrt (3))/(2) Is the student correct?If so why, if not why? Explain your answer:

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LawrenceMaster · Tutor for 5 years

Answer

The student is not correct. The correct coefficients for the quadratic equation are \(a = 2\), \(b = 9\), and \(c = 30\). Using these values in the quadratic formula shows that the equation has no real solutions, as the discriminant is negative.

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## Step 1: Identifying the Coefficients<br />The quadratic equation in standard form is \(ax^2 + bx + c = 0\). <br />In the given equation \(2x^2 + 30 = -9x\), we need to rearrange it to fit this standard form.<br />Therefore, the equation becomes \(2x^2 + 9x + 30 = 0\).<br />Here, \(a = 2\), \(b = 9\) (not 30 as the student identified), and \(c = 30\).<br /><br />## Step 2: Applying the Quadratic Formula<br />The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).<br />Substituting the correct values of \(a\), \(b\), and \(c\) gives us:<br />\[x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot 30}}{2 \cdot 2}\]<br />\[x = \frac{-9 \pm \sqrt{81 - 240}}{4}\]<br />\[x = \frac{-9 \pm \sqrt{-159}}{4}\]<br /><br />Since the discriminant (\(b^2 - 4ac\)) is negative (\(-159\)), this means the equation has no real solutions. The student's solution is incorrect as they used the wrong coefficient for \(b\) and incorrectly calculated the discriminant and the solutions.<br /><br />## Step 3: Conclusion<br />The student made a fundamental error in identifying the coefficient \(b\) and thus proceeded with incorrect calculations. The correct approach shows that the equation has no real solutions, contrary to the student's conclusion of real solutions.
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