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2 years ago

Check all of the correct solutions to this problem

Mathematics

1 answer:

MA_775_DIABLO [31]2 years ago

8 0

To solve a quadratic equation $ax^2+bx+c=0$ we can use the quadratic formula:

$x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In your case, the equation is identified by $a=c=1,\ b=-5$

So, the solving formula yields

$x_{1,2} = \cfrac{5\pm\sqrt{25-4}}{2} = \cfrac{5\pm\sqrt{21}}{2}$

So, the two solutions are derived by choosing the plus sign once, and the minus sign the other time:

$x_{1} = \cfrac{5+\sqrt{21}}{2},\quad x_{2} = \cfrac{5-\sqrt{21}}{2}$

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Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/

SVEN [57.7K]

Given $p,q>0$ and $\log_9p=\log_{12}q=\log_{16}(p+q)=x$ (say).

Then,

$p=9^x\\ q=12^x\\ p+q=16^x$

From the above 3 equations,

$\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\$

From the equations, we get

$\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}$

Since $p,q>0$ , the negative value is rejected.

$\frac{q}{p}=\frac{1 + \sqrt{5}}{2}$

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<h3>What are the 3 sides of a triangle?</h3>

The hypotenuse of a right triangle is its longest side; its "opposite" side is the one that faces the angle in question; and its "adjacent" side is the one that faces it. We use special terminology to define the sides of right triangles.

We use special terminology to define the sides of right triangles.

The side opposite the right angle is always the hypotenuse of a right triangle. It is the longest side in a right triangle.

The opposing and neighboring sides are the other two sides. The labels on these sides relate to an angle.

One side is across from another at a particular angle.

To learn more about sides of a triangle visit:

brainly.com/question/9449776

#SPJ4

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