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someone help

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For what value of the constant k does the quadratic $$2x^2 - 5x + k$$ have a double root?

Guest Dec 27, 2017
edited by Guest  Dec 27, 2017

#3
+1873
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Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, $$b^2-4ac$$ . The discriminant, if equal to zero, results in a quadratic with a "double root."

 $$b^2-4ac=0$$ Plug in the given values of a, b, and c of the quadratic. $$(-5)^2-4(2)(k)=0$$ Now, solve for k. $$25-8k=0$$ Subtract 25 from both sides. $$-8k=-25$$ $$k=\frac{25}{8}$$
TheXSquaredFactor  Dec 27, 2017
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#1
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I understood until the equation part but what do u mean by double root?

Rauhan  Dec 27, 2017
#2
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or do u mean square root/ just (x2). (x = any number/ coefficients etc)

Rauhan  Dec 27, 2017
#5
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a repeated root

Guest Dec 27, 2017
#3
+1873
+2

Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, $$b^2-4ac$$ . The discriminant, if equal to zero, results in a quadratic with a "double root."

 $$b^2-4ac=0$$ Plug in the given values of a, b, and c of the quadratic. $$(-5)^2-4(2)(k)=0$$ Now, solve for k. $$25-8k=0$$ Subtract 25 from both sides. $$-8k=-25$$ $$k=\frac{25}{8}$$
TheXSquaredFactor  Dec 27, 2017
#4
+502
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does he mean like finding the value of k and then putting the answer as $$\frac{5^2}{8}$$

Rauhan  Dec 27, 2017
#6
+1873
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I do not believe so; there is no reason not to simplify the fraction completely.

TheXSquaredFactor  Dec 28, 2017

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