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MathematicsPrecalculusQuadratic Equations → Finding Intercepts

Instructions for problem set  Find x-intercepts of the given quadratic equation

Questions


\(y = x^2 + 5 x + 4\)

\(y = x^2 -x\)

\(y = 4 x^2 -10 x + 4\)

\(y = 3 x^2 + 7 x -6\)

\(y = x^2 + 4 x + 3\)

\(y = x^2 + 7 x + 10\)

\(y = x^2 + 3 x -10\)

\(y = x^2 -3 x -4\)

\(y = x^2 + 3 x -4\)

\(y = 4 x^2 -7 x + 3\)

\(y = x^2 + 7 x + 6\)

\(y = x^2 -25\)

\(y = 3 x^2 + 10 x + 3\)

\(y = x^2 + 6 x\)

\(y = x^2 -2 x -8\)

\(y = x^2 -3 x\)

\(y = x^2 -2 x -8\)

\(y = x^2 -x -6\)

\(y = 3 x^2 -3\)

\(y = x^2 -2 x + 1\)

\(y = x^2 -11 x + 30\)

\(y = 3 x^2 -9 x + 6\)

\(y = 3 x^2 -3 x -6\)

\(y = 2 x^2 -5 x -3\)

\(y = x^2 -2 x -3\)

\(y = x^2 -x -20\)

\(y = x^2 + 7 x + 6\)

\(y = 2 x^2 -3 x -2\)

Answers


\(y = x^2 + 5 x + 4 \) ⇒
\(-1, -4\)

\(y = x^2 -x \) ⇒
\(1, 0\)

\(y = 4 x^2 -10 x + 4 \) ⇒
\(1 / 2, 2\)

\(y = 3 x^2 + 7 x -6 \) ⇒
\(2 / 3, -3\)

\(y = x^2 + 4 x + 3 \) ⇒
\(-3, -1\)

\(y = x^2 + 7 x + 10 \) ⇒
\(-5, -2\)

\(y = x^2 + 3 x -10 \) ⇒
\(-5, 2\)

\(y = x^2 -3 x -4 \) ⇒
\(4, -1\)

\(y = x^2 + 3 x -4 \) ⇒
\(-4, 1\)

\(y = 4 x^2 -7 x + 3 \) ⇒
\(3 / 4, 1\)

\(y = x^2 + 7 x + 6 \) ⇒
\(-1, -6\)

\(y = x^2 -25 \) ⇒
\(-5, 5\)

\(y = 3 x^2 + 10 x + 3 \) ⇒
\(-1 / 3, -3\)

\(y = x^2 + 6 x \) ⇒
\(0, -6\)

\(y = x^2 -2 x -8 \) ⇒
\(4, -2\)

\(y = x^2 -3 x \) ⇒
\(0, 3\)

\(y = x^2 -2 x -8 \) ⇒
\(-2, 4\)

\(y = x^2 -x -6 \) ⇒
\(3, -2\)

\(y = 3 x^2 -3 \) ⇒
\(-1, 1\)

\(y = x^2 -2 x + 1 \) ⇒
\(1, 1\)

\(y = x^2 -11 x + 30 \) ⇒
\(5, 6\)

\(y = 3 x^2 -9 x + 6 \) ⇒
\(1, 2\)

\(y = 3 x^2 -3 x -6 \) ⇒
\(-1, 2\)

\(y = 2 x^2 -5 x -3 \) ⇒
\(-1 / 2, 3\)

\(y = x^2 -2 x -3 \) ⇒
\(3, -1\)

\(y = x^2 -x -20 \) ⇒
\(-4, 5\)

\(y = x^2 + 7 x + 6 \) ⇒
\(-6, -1\)

\(y = 2 x^2 -3 x -2 \) ⇒
\(-1 / 2, 2\)
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