Math Problems

MathematicsPrecalculusQuadratic Equations → Finding Intercepts

Find x-intercepts of the given quadratic equation

## Questions

 $$y = 2 x^2 + 5 x + 3$$ $$y = 4 x^2 -14 x + 6$$ $$y = 3 x^2 -3 x$$ $$y = x^2 + 3 x -10$$ $$y = 3 x^2 + 10 x + 3$$ $$y = x^2 + x -20$$ $$y = x^2 -8 x + 12$$ $$y = x^2 -6 x$$ $$y = 3 x^2$$ $$y = 3 x^2 + 6 x$$ $$y = 4 x^2 + 9 x -9$$ $$y = x^2 + 2 x -15$$ $$y = x^2 + 2 x + 1$$ $$y = 3 x^2 + 12 x + 9$$ $$y = 3 x^2 + 9 x$$ $$y = x^2 + 6 x + 9$$ $$y = x^2 -1$$ $$y = x^2 -16$$ $$y = x^2 -6 x + 9$$ $$y = 3 x^2 + 5 x -2$$ $$y = 4 x^2 -7 x -2$$ $$y = x^2 -3 x -10$$ $$y = 2 x^2 -3 x + 1$$ $$y = 4 x^2 + x -3$$ $$y = x^2 + x$$ $$y = x^2 -3 x$$ $$y = 3 x^2 + 3 x$$ $$y = x^2 + 5 x$$

## Answers

 $$y = 2 x^2 + 5 x + 3$$ ⇒ $$-3 / 2, -1$$ $$y = 4 x^2 -14 x + 6$$ ⇒ $$1 / 2, 3$$ $$y = 3 x^2 -3 x$$ ⇒ $$0, 1$$ $$y = x^2 + 3 x -10$$ ⇒ $$2, -5$$ $$y = 3 x^2 + 10 x + 3$$ ⇒ $$-1 / 3, -3$$ $$y = x^2 + x -20$$ ⇒ $$-5, 4$$ $$y = x^2 -8 x + 12$$ ⇒ $$6, 2$$ $$y = x^2 -6 x$$ ⇒ $$6, 0$$ $$y = 3 x^2$$ ⇒ $$0, 0$$ $$y = 3 x^2 + 6 x$$ ⇒ $$0, -2$$ $$y = 4 x^2 + 9 x -9$$ ⇒ $$3 / 4, -3$$ $$y = x^2 + 2 x -15$$ ⇒ $$3, -5$$ $$y = x^2 + 2 x + 1$$ ⇒ $$-1, -1$$ $$y = 3 x^2 + 12 x + 9$$ ⇒ $$-1, -3$$ $$y = 3 x^2 + 9 x$$ ⇒ $$0, -3$$ $$y = x^2 + 6 x + 9$$ ⇒ $$-3, -3$$ $$y = x^2 -1$$ ⇒ $$1, -1$$ $$y = x^2 -16$$ ⇒ $$4, -4$$ $$y = x^2 -6 x + 9$$ ⇒ $$3, 3$$ $$y = 3 x^2 + 5 x -2$$ ⇒ $$1 / 3, -2$$ $$y = 4 x^2 -7 x -2$$ ⇒ $$-1 / 4, 2$$ $$y = x^2 -3 x -10$$ ⇒ $$-2, 5$$ $$y = 2 x^2 -3 x + 1$$ ⇒ $$1 / 2, 1$$ $$y = 4 x^2 + x -3$$ ⇒ $$3 / 4, -1$$ $$y = x^2 + x$$ ⇒ $$-1, 0$$ $$y = x^2 -3 x$$ ⇒ $$0, 3$$ $$y = 3 x^2 + 3 x$$ ⇒ $$0, -1$$ $$y = x^2 + 5 x$$ ⇒ $$0, -5$$
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