Question

find the quadratic equation whose roots are sin 30 and tan 45​

Answer 1

For any two zeroes a and b, the equation will be

x² - (a+b)x + ab

In this case a = sin 30 = ½

b = tan 45 = 1/√2

Hence the equation will be,

${x}^{2} - ( \frac{1}{2} + \frac{1}{ \sqrt{2} } )x + \frac{1}{2} \times \frac{1}{ \sqrt{2} } \\ {x}^{2} - \frac{( \sqrt{2} + 2) x}{2 \sqrt{2} } + \frac{1}{2 \sqrt{2} } \\ \\ multiplying \: all \: the \: terms \: by \: 2 \sqrt{2} \\ 2 \sqrt{2} {x}^{2} - (2 + \sqrt{2} )x + 1$

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Answer 2

Answer: The quadratic equation whose roots are sin 30° and tan 45° is:

$x = (4c(tan 45°) / 3 +/- sqrt((4c(tan 45°) / 3)^2 - 4(4c/3)(c / 3/4))) / (8c/3)$

Step-by-step explanation

To find the quadratic equation whose roots are sin 30° and tan 45°, we can use the fact that the roots of a quadratic equation of the form

$ax^2 + bx + c = 0$

are given by the formula

$x = (-b +/- sqrt(b^2 - 4ac)) / (2a).$

So, we want to find values for a, b, and c such that the roots of the equation are sin 30° and tan 45°.

Let's let x = sin 30° be one of the roots, so we have:

$ax^2 + bx + c = 0$

$a(sin 30°)^2 + b(sin 30°) + c = 0$

Next, let's let y = tan 45° be the other root, so we have:

$ay^2 + by + c = 0$

$a(tan 45°)^2 + b(tan 45°) + c = 0$

We can now solve this system of equations to find the values of a, b, and c.

Substituting the value of sin 30° for x and the value of tan 45° for y, we get:

$a(sin 30°)^2 + b(sin 30°) + c = 0$

$a(tan 45°)^2 + b(tan 45°) + c = 0$

Solving for a, we get:

$a = c / (sin 30°)^2 = c / (cos^2 30°)$

$= c / (1 - sin^2 30°)$

$= c / (1 - (1/2)^2)$

$= c / (1 - 1/4)$

$= c / 3/4$

$= 4c / 3$

Substituting this expression for a into the second equation, we get:

$(4c/3)(tan 45°)^2 + b(tan 45°) + c = 0$

Solving for b, we get:

$b = -4c(tan 45°) / 3$

Finally, substituting these expressions for a and b into the general quadratic formula, we get:

$x = (-b +/- sqrt(b^2 - 4ac)) / (2a)$

$= (-(-4c(tan 45°) / 3) +/- sqrt((-4c(tan 45°) / 3)^2 - 4(4c/3)(c / 3/4))) / (2(4c/3))$

$= (4c(tan 45°) / 3 +/- sqrt((4c(tan 45°) / 3)^2 - 4(4c/3)(c / 3/4))) / (8c/3)$

Thus, the quadratic equation whose roots are sin 30° and tan 45° is:

$x = (4c(tan 45°) / 3 +/- sqrt((4c(tan 45°) / 3)^2 - 4(4c/3)(c / 3/4))) / (8c/3)$

For more question on quadratic equation-https://brainly.com/question/30098550

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