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3 years ago

Using the quadratic formula, what is the solution to 2x^2-5x-6=0 ?

1 answer:

4 0

$\bf \textit{quadratic formula}\\

\begin{array}{lcclll}
2x^2&-5x&-6&=0\\
\uparrow &\uparrow &\uparrow \\
a&b&c
\end{array}
\qquad \qquad 
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}\\\\
-----------------------------\\\\
x=\cfrac{-(-5)\pm\sqrt{(-5)^2-4(2)(-6)}}{2(2)}\implies x=\cfrac{5\pm\sqrt{25+48}}{4}
\\\\\\
x=\cfrac{5\pm\sqrt{73}}{4}\implies x=
\begin{cases}
\cfrac{5+\sqrt{73}}{4}\\\\
\cfrac{5-\sqrt{73}}{4}
\end{cases}$

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Given:

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For $x=0$ ,

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Divide both sides by 2.

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The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

$(x,y)\to (3x,3y)$

Using this rule, we get

$A(0,5)\to A'(3(0),3(5))$

$A(0,5)\to A'(0,15)$

Similarly,

$B(2,2)\to B'(3(2),3(2))$

$B(2,2)\to B'(6,6)$

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

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$y-15=\dfrac{6-15}{6-0}(x-0)$

$y-15=\dfrac{-9}{6}(x)$

$y-15=\dfrac{-3}{2}x$

Multiply both sides by 2.

$2(y-15)=-3x$

$2y-30=-3x$

$2y+3x=30$

Therefore, the equation of the line after the dilation is $2y+3x=30$ .

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