Algebra · real student question

Solve 4x² + 16x + 7 = 0.

Question

Solve

4x2+16x+7=04x^{2}+16x+7=0

Step-by-step solution

  1. Compute the discriminant first. With a=4a=4, b=16b=16, c=7c=7,

    b24ac=1624(4)(7)=256112=144b^{2}-4ac=16^{2}-4(4)(7)=256-112=144

    A positive perfect square means two distinct rational roots — so both the quadratic formula and factoring will work, and the formula is the more mechanical route.

  2. Apply the quadratic formula.

    x=b±b24ac2a=16±1448=16±128x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-16\pm\sqrt{144}}{8}=\frac{-16\pm 12}{8}

  3. Evaluate the two branches.

    x=16+128=48=12,x=16128=288=72x=\frac{-16+12}{8}=\frac{-4}{8}=-\frac12,\qquad x=\frac{-16-12}{8}=\frac{-28}{8}=-\frac72

    x=12 or x=72\boxed{x=-\tfrac12\ \text{or}\ x=-\tfrac72}

  4. Cross-check by factoring. Because the discriminant was a perfect square, the quadratic factors over the rationals:

    4x2+16x+7=(2x+1)(2x+7)4x^{2}+16x+7=(2x+1)(2x+7)

    Expanding: 4x2+14x+2x+7=4x2+16x+74x^{2}+14x+2x+7=4x^{2}+16x+7 ✓, and setting each bracket to zero gives the same two roots.

  5. Check with Vieta's formulas. The roots must sum to b/a=4-b/a=-4 and multiply to c/a=74c/a=\tfrac74: 1272=4-\tfrac12-\tfrac72=-4 ✓ and (12)(72)=74\left(-\tfrac12\right)\left(-\tfrac72\right)=\tfrac74 ✓. Both signs being negative is also expected, since bb and cc are both positive with a>0a>0.

Answer

x=12 or x=72x=-\dfrac{1}{2}\ \text{or}\ x=-\dfrac{7}{2}

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