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3 years ago

How can I factor 5x^2 + 26x - 24 = 0 using the completing the square method.

Mathematics

2 answers:

Tema [17]3 years ago

5 0

The answer is -8 with using the completing the square method

kifflom [539]3 years ago

3 0

<h2><u>QUADRATIC EQUATION</u></h2>

<h2> $\mathbb{ANSWER:}$ </h2>

$\bold{x = 0.8 \: } \: \sf \: {\color{grey}or} \: \: \: \bold{ x= \frac{4}{5} } \\$

$\bold{x = - 6}$

— — — — — — — — — —

<h3>Step-by-step explanation:</h3>

<u>How can </u><u>we</u><u> factor 5x^2 + 26x - 24 = 0 using the completing the square </u><u>method?</u>

Let's solve your equation step-by-step.

$\bold{Given \: Equation: \color{brown} 5x²+26x-24=0}$

First, add 24 to both sides.

$\bold{5x²+26x-24 - \purple{ 24} = 0 + \purple{24}}$

$\bold{ \implies \: 5x²+26x = 24 }$

Since the coefficient of 5x² is 5, divide both sides by 5.

$\bold{ \frac{5 {x}^{2} + 26x }{5} = \frac{24}{5} } \\$

$\bold{ \implies \: {x}^{2} + \frac{26}{5} x = \frac{24}{5} } \\$

The coefficient of 26/5x is 26/5. So, let b=26/5.

Then we need to add (b/2)²=169/25 to both sides to complete the square.

Add 169/25 to both sides.

$\bold{ {x}^{2} + \frac{26}{5} x + \frac{ \purple{169}}{ \purple{25}}= \frac{24}{5} + \frac{ \purple{169}}{ \purple{25}} } \\$

$\bold{ \implies \:\bold{ {x}^{2} + \frac{26}{5} x + \frac{ 169}{ {25}}= \frac{289}{25} } } \\$

Factor the left side.

$\bold{(x + \frac{13}{5} ) {}^{2} = \frac{289}{25} } \\$

Take square root.

$\bold{x + \frac{13}{5} = ± \: \sqrt{ \frac{289}{25} }} \\$

Then, add (-13)/5 to both sides.

$\bold{x + \frac{13}{5} + \frac{\purple{ - 13} }{\purple{ 5}} = \frac{{ - 13} }{{ 5}} ± \: \sqrt{ \frac{289}{25}}} \\$

$\bold{ \: x = \frac{ - 13}{5} ± \sqrt{ \frac{289}{25} } } \\$

$\implies \: \underline{ \boxed{ \bold{ \frac{4}{5} } \sf \: \: or \: \: \bold{ x = - 6}}} \\$

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Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

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Answer:

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3 years ago

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malfutka [58]

PART A

The given equation is

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In order to find the maximum height, we write the function in the vertex form.

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We add and subtract

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to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

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This finally simplifies to,

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The vertex of this function is

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The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

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This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

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