What is the least five-digit even number with no repeated digits?

Solve using the quadratic formula. Show all work. Write each solution in simplest form. No decimals.

r-ruslan [8.4K]

Given:

The given quadratic equation is:

$12a^2+9a+7=0$

To find:

The solution for the given quadratic equation in simplest form by using the quadratic formula.

Solution:

Consider a quadratic equation is defined as $ax^2+bx+c=0$ , then the quadratic formula is:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

The given quadratic equation is:

$12a^2+9a+7=0$

We have, $a=12,b=9,c=7$ . Using the quadratic formula, we get

$x=\dfrac{-(9)\pm \sqrt{(9)^2-4(12)(7)}}{2(12)}$

$x=\dfrac{-9\pm \sqrt{81-336}}{24}$

$x=\dfrac{-9\pm \sqrt{-255}}{24}$

$x=\dfrac{-9\pm i\sqrt{255}}{24}$ $[\because \sqrt{-a}=i\sqrt{a}]$

The solutions of the given equations are $x=\dfrac{-9-i\sqrt{255}}{24}$ and $x=\dfrac{-9+i\sqrt{255}}{24}$ .

Hence, the correct option is B.

3 0

3 years ago

The function h is defined by h(y)=2 y-7 . what is the value of h(8)

gtnhenbr [62]

I believe the answer is 2

7 0

4 years ago

5x+25=75 pls help appreciated

Dafna11 [192]

Answer:

x=10

Step-by-step explanation:

I guess solve for x

5x=50

x=10

3 0

3 years ago

Read 2 more answers

When constructing an equilateral triangle by hand, which step comes after constructing a circle?

KonstantinChe [14]

Answer:

set compass to the radius of the circle

Step-by-step explanation: when constructing an equilateral triangle by hand . circle constructing by set compass to the radius of the circle and rotate it and come to the initial point

4 0

3 years ago

Read 2 more answers

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

nignag [31]

Standard reduction of order procedure: suppose there is a second solution of the form $y_2(x)=v(x)y_1(x)$ , which has derivatives

$y_2=vx^4$
${y_2}'=v'x^4+4vx^3$
${y_2}''=v''x^4+8v'x^3+12vx^2$

Substitute these terms into the ODE:

$x^2(v''x^4+8v'x^3+12vx^2)-7x(v'x^4+4vx^3)+16vx^4=0$
$v''x^6+8v'x^5+12vx^4-7v'x^5-28vx^4+16vx^4=0$
$v''x^6+v'x^5=0$

and replacing $v'=w$ , we have an ODE linear in $w$ :

$w'x^6+wx^5=0$

Divide both sides by $x^5$ , giving

$w'x+w=0$

and noting that the left hand side is a derivative of a product, namely

$\dfrac{\mathrm d}{\mathrm dx}[wx]=0$

we can then integrate both sides to obtain

$wx=C_1$
$w=\dfrac{C_1}x$

Solve for $v$ :

$v'=\dfrac{C_1}x$
$v=C_1\ln|x|+C_2$

Now

$y=C_1x^4\ln|x|+C_2x^4$

where the second term is already accounted for by $y_1$ , which means $y_2=x^4\ln x$ , and the above is the general solution for the ODE.

4 0

4 years ago