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If α and β are the zeros of the quadratic polynomial f(x) = 6x²+x-2, then the value of

1. α²+β²
2. 1/α+1/β​

Can Someone Answer This Quick Thank You<3

Sagot :

semsee45

Answer:

Given function:

[tex]f(x)=6x^2+x-2[/tex]

To find the zeros of the function, set the function to zero and factor:

[tex]\implies 6x^2+x-2=0[/tex]

[tex]\implies 6x^2+4x-3x-2=0[/tex]

[tex]\implies 2x(3x+2)-1(3x+2)=0[/tex]

[tex]\implies (2x-1)(3x+2)=0[/tex]

Therefore, the zeros are:

[tex]\implies (2x-1)=0 \implies x=\dfrac{1}{2}[/tex]

[tex]\implies (3x+2)=0 \implies x=-\dfrac{2}{3}[/tex]

If α and β are the zeros of the function:

  • [tex]\textsf{Let } \alpha=\dfrac{1}{2}[/tex]
  • [tex]\textsf{Let } \beta=-\dfrac{2}{3}[/tex]

Question 1

[tex]\begin{aligned}\implies \alpha^2+\beta^2 & =\left(\dfrac{1}{2}\right)^2+\left(-\dfrac{2}{3}\right)^2\\\\& = \dfrac{1}{4}+\dfrac{4}{9}\\\\& = \dfrac{9}{36}+\dfrac{16}{36}\\\\& = \dfrac{25}{36}\end{aligned}[/tex]

Question 2

[tex]\begin{aligned}\implies \dfrac{1}{\alpha}+\dfrac{1}{\beta} & = \dfrac{1}{\frac{1}{2}}+\dfrac{1}{-\frac{2}{3}}\\\\& = 1 \times \dfrac{2}{1}+1 \times -\dfrac{3}{2}\\\\& = 2 - \dfrac{3}{2}\\\\& = \dfrac{1}{2}\end{aligned}[/tex]

loneguy

Answer:

[tex]1) ~\alpha^2+\beta^2 = \dfrac{25}{36}\\\\\\2)~\dfrac 1 {\alpha} + \dfrac 1{\beta} = \dfrac 12[/tex]

Step-by-step explanation:

[tex]\text{Given that,}\\\\f(x) = 6x^2 +x -2~ \text{and the roots are}~ \alpha, \beta\\\\\text{Now,}\\\\\alpha + \beta = -\dfrac{b}{a} = -\dfrac{1}{6}~~~~~;[\text{Compare with the standard form}~ ax^2 +b x + c = 0]\\\\\alpha \beta = \dfrac ca = -\dfrac2 6 = - \dfrac 13\\\\\\\textbf{1)}\\\\\alpha^2 +\beta^2\\\\\\=\left(\alpha +\beta \right)^2 - 2 \alpha \beta \\\\\\=\left( -\dfrac 16 \right)^2 -2 \left(- \dfrac 13 \right)\\\\\\=\dfrac{1}{36}+\dfrac 23\\\\\\=\dfrac{25}{36}[/tex]

[tex]\textbf{2)}\\\\\dfrac 1{\alpha} + \dfrac 1{\beta} \\\\\\=\dfrac{\alpha + \beta}{\alpha \beta }\\\\\\=\dfrac{-\tfrac16}{-\tfrac 13}\\\\\\=\dfrac{1}{6} \times 3\\\\\\=\dfrac{1}{2}[/tex]

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