Step-by-step explanation:
Since, roots are - 3 & 2
Therefore, (x + 3) & (x - 2) would be factors.
Leading coefficient is 5
Hence, required quadratic equation is:
[tex]5(x + 3)(x - 2) = 0 \\ \therefore \: 5( {x}^{2} - 2x + 3x - 6) = 0\\ \therefore \: 5( {x}^{2} + x - 6) = 0\\ \therefore \: 5 {x}^{2} + 5x - 30 = 0\: \\ is \: the \: required \: quadratic \: equation.[/tex]
The quadratic equation be [tex]$5 x^{2}+5 x-30=0[/tex]
(x + 3) and (x - 2) would be factors of the quadratic equation.
Any equation having one term in which the unknown exists squared and no term in which it exists raised to a higher power solve for x in the quadratic equation [tex]x^{2} + 4x + 4 = 0.[/tex]
We define a quadratic equation as an equation of degree 2, indicating that the highest exponent of this function exists 2.
Given:
the roots exist -3 and 2.
The leading coefficient exists 5.
To find:
the quadratic equation.
Hence, the required quadratic equation exists:
[tex]5(x + 3)(x - 2) = 0[/tex]
simplifying the above equation, we get
[tex]$5\left(x^{2}-2 x+3 x-6\right)=0 \\[/tex]
[tex]$ 5\left(x^{2}+x-6\right)=0 \\[/tex]
[tex]$5 x^{2}+5 x-30=0[/tex]
Therefore, [tex]$5 x^{2}+5 x-30=0[/tex] exists the required quadratic equation.
To learn more about quadratic equations
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