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Find an equation of a quadratic function whose zeroes are 3/2 and -1/3?
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Sagot :

[tex]x = \frac{3}{2} \: and \: x = - \frac{1}{3} \: are \: two \: real \: district \: solutions \: for \: quadratic \: equations, \: which \: means \: that \: x - \frac{3}{2} \: and \: x + \frac{1}{3} \: are \: the \: factors \: of \: the \: quadratic \: equation.[/tex]

[tex](x - \frac{3}{2} )(x + \frac{1}{2} ) = 0[/tex]

[tex]expand \: (x - \frac{3}{2} )(x + \frac{1}{3} ) \: using \: the \: foil \: method.[/tex]

[tex] x \times x + x( \frac{1}{2} - \frac{3}{2}x - \frac{3}{2} \times \frac{1}{23} = 0[/tex]

Simplify and combine like terms.

[tex] \frac{ {x}^{2} \times 6 + x \times 2 - 3x \times 3 - 3 }{6} = 0[/tex]

Simplify the numerator.

[tex] \frac{(3x + 1)(2x - 3)}{6} = 0[/tex]

Expand ( 3x + 1 ) ( 2x − 3 ) using the FOIL Method.

[tex] \frac{3x(2x) + 3x \times - 3 + 1(2x) + \times - 3}{6} = 0[/tex]

Simplify and combine like terms.

[tex] \frac{6 {x}^{2} - 7x - 3 }{6} = 0[/tex]

[tex] \frac{6 {x}^{2} - 7x}{6} + \frac{ - 3}{6} = 0[/tex]

Cancel the common factor of 6.

[tex] {x}^{2} + \frac{ - 7x}{6} + \frac{ - 3}{6} =0[/tex]

Move the negative in front of the fraction.

[tex] {x}^{2} - \frac{ - 7x}{6} + \frac{ - 3}{6} = 0[/tex]

Cancel the common factor of -3 and 6.

[tex] {x}^{2} - \frac{7x}{6} + \frac{ - 1}{2} = 0[/tex]

Move the negative in front of the fraction.

[tex] {x}^{2} - \frac{ - 7}{6} - \frac{1}{2} = 0[/tex]

The standard of quadratic equation using the given set of solutions {3/2,−1/3} is

[tex]y = {x}^{2} - \frac{7x}{6} - \frac{1}{2} [/tex]

[tex]y = {x}^{2} - \frac{7x}{6} - \frac{1}{2} [/tex]

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