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lukranit [14]
3 years ago
6

Determine the points of intersection of the line y = -2x + 7 and the parabola y = 2x2 + 3x - 5.

Mathematics
1 answer:
Nata [24]3 years ago
8 0

Answer:


Step-by-step explanation:

The x-values of those points can be found by equating the expressions for y, then solving for x.

... -2x +7 = 2x² +3x -5

... 2x² +5x -12 = 0 . . . . . . . . subtract the left side to put into standard form

This can be factored by looking fro two factors of 2·(-12) = -24 that add to give 5, the x-coefficient. Those factors are 8, -3.

... 2x² +8x -3x -12 = 0 . . . . rewrite the middle term

... 2x(x +4) -3(x +4) = 0 . . . factor by pairs

... (x +4)(2x -3) = 0 . . . . . . . complete factorization

... x = -4, x = 3/2 . . . . . . . . values of x that make the factors be zero

The corresponding y-values are ...

... for x = -4, y = -2(-4) +7 = 15

... for x = 3/2, y = -2(3/2) +7 = 4

The points of intersection are (-4, 15) and (3/2, 4).

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The value of x is x=\frac{-1+\sqrt{37}}{12} and x=\frac{-1-\sqrt{37}}{12}

Step-by-step explanation:

The equation is \frac{1}{x}-\frac{2}{3}=4 x

Subtracting by 4x on both sides,

\frac{1}{x}-\frac{2}{3}-4 x=0

Taking LCM,

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Multiplying by 3x on both sides,

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Dividing by (-) on both sides,

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Using quadratic formula, we can solve for x.

\begin{aligned}x &=\frac{-2 \pm \sqrt{2^{2}-4 \cdot 12 \cdot(-3)}}{2 \cdot 12} \\&=\frac{-2 \pm \sqrt{4+144}}{2 \cdot 144} \\&=\frac{-2 \pm \sqrt{148}}{24} \\&=\frac{-2 \pm 2 \sqrt{37}}{24}\end{aligned}

Taking out common term 2, we get,

\begin{array}{l}{x=\frac{-2(1 \pm \sqrt{37})}{24}} \\{x=\frac{-1 \pm \sqrt{37}}{12}}\end{array}

Thus, the value of x is  x=\frac{-1+\sqrt{37}}{12} and x=\frac{-1-\sqrt{37}}{12}

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