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3 years ago

Find the solution to the equation below. 2x^2+3x-20=0

Mathematics

1 answer:

mariarad [96]3 years ago

8 0

Answer:

x = -4, 5/2

Step-by-step explanation:

A quadratic can be solved may ways, including graphing, factoring, and the quadratic formula. You can also check possible answers by making use of the relationships between solutions and the coefficients.

A graph is attached. It shows the solutions to be -4 and 5/2.

When factored, the equation becomes ...

(2x -5)(x +4) = 0 . . . . . has solutions x=-4, x=5/2 (these make the factors zero)

Using the quadratic formula, the solutions of ax^2 +bx +c = 0 are found from ...

x = (-b±√(b²-4ac))/(2a)

x = (-3±√(3²-4(2)(-20))/(2(2)) = (-3±√169)/4 = {-16, +10}/4

x = {-4, 5/2}

For ax^2 +bx +c = 0, the solutions must satisfy ...

product of solutions is c/a = -20/2 = -10

Only the first and last choices have this product.

sum of solutions is -b/a = -3/2

Only the first choice (-4, 5/2) has this sum.

You might be interested in

The side length of a square is represented by the expression 2x + 5. Which expression represents the difference between the area

musickatia [10]

Answer:

$4x^2 + 22x + 15$

Step-by-step explanation:

The side length of a square is represented by the expression 2x + 5.

The area of a square is given as:

$A = a^2$

where a = length of side of the square

The area of the square is therefore:

$A = (2x + 5)^2\\\\A = (2x + 5)(2x + 5)\\\\A = 4x^2 + 20x + 25$

The perimeter of a square is given as:

$P = 4a$

The perimeter of the square is therefore:

$P = 4(2x + 5) \\\\P = 8x + 10$

The difference between the area of the square and the perimeter of the square is:

$4x^2 + 30x + 25 - (8x + 10)\\\\4x^2 + 30x + 25 - 8x - 10\\\\4x^2 + 30x - 8x + 25 -10\\\\4x^2 + 22x + 15$

The expression that represents the difference between the area and the perimeter of the square is:

$4x^2 + 22x + 15$

7 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :