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3 years ago

Please help me with this problem

Mathematics

1 answer:

Rom4ik [11]3 years ago

6 0

y=x^2 + x - 2

x + y = 1

Replace y in the second equation:

x + x^2 + x -2 = 1

Simplify:

x^2 + 2x -2 = 1

Subtract 1 from both sides:

x^2 + 2x -3 = 0

Factor:

(x-1) (x+3) = 0

Solve for both x's:

x = 1 and x = -3

Now replace x in the second equation and solve for y using both x values:

1 + y = 1, y = 0

-3 + y = 1, y = 4

Now you have (1,0) and (-3,4) as solutions for (x,y)

XY = x times y:

1 x 0 = 0

-3 x 4 = -12

The answer would be -12

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What is the solution to -x2 + 5x - 3

CaHeK987 [17]

The solution to the equation is $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Explanation:

Given that the equation is $-x^2+5x-3=0$

We need to determine the solution of the equation.

The solution of the equation can be determined using the quadratic formula.

The quadratic formula is given by

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

Hence, from the equation, we have,

$a=-1,\:b=5,\:c=-3$

Substituting these values in the quadratic formula, we get,

$x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$

Simplifying, we get,

$x=\frac{-5\pm \sqrt{25-12}}{-2}$

Simplifying the terms within the root, we get,

$x=\frac{-5\pm \sqrt{13}}{-2}$

Thus, the roots of the equation are $x=\frac{-5+ \sqrt{13}}{-2}$ and $x=\frac{-5- \sqrt{13}}{-2}$

Taking out the negative sign, we get,

$x=\frac{-(5- \sqrt{13})}{-2}$ and $x=\frac{-(5+ \sqrt{13})}{-2}$

Cancelling the negative sign, we get,

$x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Thus, the solutions of the equation are $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

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Please help me with this problem​

Please help me with this problem