What is the quadratic equation that has roots of ,-3 And 2/5

Mathematics

1 answer:

prohojiy [21]2 years ago

6 0

Answer: 5x² + 13x - 6 = 0

Step-by-step explanation:

Represent the roots with X i.e.

X = -3 ; X = 2/5

Moving the values of X over the equator we have;

X + 3 = 0 ; 5X - 2 = 0

Equating both equations to zero we can obtain;

(X + 3)(5X - 2) = 0

Opening the brackets we get;

5x² - 2x + 15x - 6 = 0

The result of the above equation gives;

5x² + 13x - 6 = 0

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write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

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We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.