Solving Quadratic Equations by Completing the Square Practice

Use the normalized linear coefficient and square after halving.

Take half of the linear coefficient and square it: \(\left(\frac{6}{2}\right)^2=9\).

Use the normalized linear coefficient and square after halving.

Take half of the linear coefficient and square it: \(\left(\frac{-8}{2}\right)^2=16\).

Use the normalized linear coefficient and square after halving.

Take half of the linear coefficient and square it: \(\left(\frac{10}{2}\right)^2=25\).

Use the normalized linear coefficient and square after halving.

Take half of the linear coefficient and square it: \(\left(\frac{-14}{2}\right)^2=49\).

Use the normalized linear coefficient and square after halving.

Take half of the linear coefficient and square it: \(\left(\frac{18}{2}\right)^2=81\).

After creating the square, retain both plus and minus square-root branches.

Add \(\left(\frac{6}2\right)^2=9\) to both sides. The perfect-square equation produces both branches, \(x=-7\) and \(x=1\), and each satisfies the original equation.

After creating the square, retain both plus and minus square-root branches.

Add \(\left(\frac{-8}2\right)^2=16\) to both sides. The perfect-square equation produces both branches, \(x=-1\) and \(x=9\), and each satisfies the original equation.

After creating the square, retain both plus and minus square-root branches.

Add \(\left(\frac{10}2\right)^2=25\) to both sides. The perfect-square equation produces both branches, \(x=-11\) and \(x=1\), and each satisfies the original equation.

After creating the square, retain both plus and minus square-root branches.

Add \(\left(\frac{-4}2\right)^2=4\) to both sides. The perfect-square equation produces both branches, \(x=-2\) and \(x=6\), and each satisfies the original equation.

After creating the square, retain both plus and minus square-root branches.

Add \(\left(\frac{2}2\right)^2=1\) to both sides. The perfect-square equation produces both branches, \(x=-5\) and \(x=3\), and each satisfies the original equation.

Divide every term by the leading coefficient before computing the half-and-square term.

Divide through by \(2\), move the constant, and complete the square. The equation becomes \((x-3)^2=4\), so \(x=1\) or \(x=5\). Direct substitution verifies both roots.

Divide every term by the leading coefficient before computing the half-and-square term.

Divide through by \(3\), move the constant, and complete the square. The equation becomes \((x+2)^2=9\), so \(x=-5\) or \(x=1\). Direct substitution verifies both roots.

Divide every term by the leading coefficient before computing the half-and-square term.

Divide through by \(4\), move the constant, and complete the square. The equation becomes \((x-1)^2=4\), so \(x=-1\) or \(x=3\). Direct substitution verifies both roots.

Divide every term by the leading coefficient before computing the half-and-square term.

Divide through by \(5\), move the constant, and complete the square. The equation becomes \((x+3)^2=1\), so \(x=-4\) or \(x=-2\). Direct substitution verifies both roots.

Divide every term by the leading coefficient before computing the half-and-square term.

Divide through by \(6\), move the constant, and complete the square. The equation becomes \((x-4)^2=4\), so \(x=2\) or \(x=6\). Direct substitution verifies both roots.

Do not convert an exact radical to a decimal unless the question requests an approximation.

Add \(4\) to both sides to get \((x+2)^2=7\). Taking both square roots yields \(x=-2\pm\sqrt{7}\).

Do not convert an exact radical to a decimal unless the question requests an approximation.

Add \(9\) to both sides to get \((x-3)^2=11\). Taking both square roots yields \(x=3\pm\sqrt{11}\).

Do not convert an exact radical to a decimal unless the question requests an approximation.

Add \(16\) to both sides to get \((x+4)^2=11\). Taking both square roots yields \(x=-4\pm\sqrt{11}\).

Do not convert an exact radical to a decimal unless the question requests an approximation.

Add \(25\) to both sides to get \((x-5)^2=31\). Taking both square roots yields \(x=5\pm\sqrt{31}\).

Do not convert an exact radical to a decimal unless the question requests an approximation.

Add \(36\) to both sides to get \((x+6)^2=29\). Taking both square roots yields \(x=-6\pm\sqrt{29}\).

Check equivalence by expanding the proposed perfect square and preserving both sides.

Adding the square of half the linear coefficient to both sides produces \((x+4)^2=21\). Expanding the left and reversing the added constant recovers the original equation.

Check equivalence by expanding the proposed perfect square and preserving both sides.

Adding the square of half the linear coefficient to both sides produces \((x-6)^2=29\). Expanding the left and reversing the added constant recovers the original equation.

Check equivalence by expanding the proposed perfect square and preserving both sides.

Adding the square of half the linear coefficient to both sides produces \((x+7)^2=51\). Expanding the left and reversing the added constant recovers the original equation.

Check equivalence by expanding the proposed perfect square and preserving both sides.

Adding the square of half the linear coefficient to both sides produces \((x-2)^2=17\). Expanding the left and reversing the added constant recovers the original equation.

Check equivalence by expanding the proposed perfect square and preserving both sides.

Adding the square of half the linear coefficient to both sides produces \((x+8)^2=61\). Expanding the left and reversing the added constant recovers the original equation.

Expand the identity and compare corresponding coefficients.

Expanding the square gives \(x^2+6x+9\). Equal polynomials have matching constant coefficients, so \(k=9\).

Expand the identity and compare corresponding coefficients.

Expanding the square gives \(x^2-8x+16\). Equal polynomials have matching constant coefficients, so \(k=16\).

Expand the identity and compare corresponding coefficients.

Expanding the square gives \(x^2+14x+49\). Equal polynomials have matching constant coefficients, so \(k=49\).

Expand the identity and compare corresponding coefficients.

Expanding the square gives \(x^2-10x+25\). Equal polynomials have matching constant coefficients, so \(k=25\).

Expand the identity and compare corresponding coefficients.

Expanding the square gives \(x^2+18x+81\). Equal polynomials have matching constant coefficients, so \(k=81\).

Coefficient comparison is valid for polynomial identities, not merely a single input equality.

Because the equality holds for every input, corresponding linear coefficients are equal. Solving that coefficient equation gives \(8\).

Coefficient comparison is valid for polynomial identities, not merely a single input equality.

Because the equality holds for every input, corresponding linear coefficients are equal. Solving that coefficient equation gives \(7\).

Coefficient comparison is valid for polynomial identities, not merely a single input equality.

Because the equality holds for every input, corresponding linear coefficients are equal. Solving that coefficient equation gives \(-4\).

Coefficient comparison is valid for polynomial identities, not merely a single input equality.

Because the equality holds for every input, corresponding linear coefficients are equal. Solving that coefficient equation gives \(8\).

Coefficient comparison is valid for polynomial identities, not merely a single input equality.

Because the equality holds for every input, corresponding linear coefficients are equal. Solving that coefficient equation gives \(5\).

Solve first, then filter roots using the stated domain or sign condition.

Both roots solve the equation, but only \(x=1\) satisfies the additional restriction \(x>0\).

Solve first, then filter roots using the stated domain or sign condition.

Both roots solve the equation, but only \(x=-1\) satisfies the additional restriction \(x<0\).

Solve first, then filter roots using the stated domain or sign condition.

Both roots solve the equation, but only \(x=-11\) satisfies the additional restriction \(x<-5\).

Solve first, then filter roots using the stated domain or sign condition.

Both roots solve the equation, but only \(x=6\) satisfies the additional restriction \(x>4\).

Solve first, then filter roots using the stated domain or sign condition.

Both roots solve the equation, but only \(x=-5\) satisfies the additional restriction \(x<0\).

Keep fractional completion terms exact and simplify the square root only after taking both branches.

Completing the square gives a squared binomial centered at \(-\frac32\) with right side \(\frac{37}{4}\). Taking both square roots and simplifying yields \(x=\frac{-3\pm\sqrt{37}}{2}\), and substitution verifies both branches.

Keep fractional completion terms exact and simplify the square root only after taking both branches.

Completing the square gives a squared binomial centered at \(\frac52\) with right side \(\frac{33}{4}\). Taking both square roots and simplifying yields \(x=\frac{5\pm\sqrt{33}}{2}\), and substitution verifies both branches.

Keep fractional completion terms exact and simplify the square root only after taking both branches.

Completing the square gives a squared binomial centered at \(-\frac72\) with right side \(\frac{45}{4}\). Taking both square roots and simplifying yields \(x=\frac{-7\pm3\sqrt5}{2}\), and substitution verifies both branches.

Keep fractional completion terms exact and simplify the square root only after taking both branches.

Completing the square gives a squared binomial centered at \(\frac92\) with right side \(\frac{97}{4}\). Taking both square roots and simplifying yields \(x=\frac{9\pm\sqrt{97}}{2}\), and substitution verifies both branches.

Keep fractional completion terms exact and simplify the square root only after taking both branches.

Completing the square gives a squared binomial centered at \(-\frac12\) with right side \(\frac{45}{4}\). Taking both square roots and simplifying yields \(x=\frac{-1\pm3\sqrt5}{2}\), and substitution verifies both branches.

Audit normalization, balance, factor expansion, and both square-root branches.

The completion term must be the square of half the linear coefficient. Expanding the perfect square and substituting the final roots provide independent checks.

Audit normalization, balance, factor expansion, and both square-root branches.

Adding unequal amounts destroys the equation's balance. Expanding the perfect square and substituting the final roots provide independent checks.

Audit normalization, balance, factor expansion, and both square-root branches.

The squared-term coefficient must be normalized first. Expanding the perfect square and substituting the final roots provide independent checks.

Audit normalization, balance, factor expansion, and both square-root branches.

Both plus and minus branches are required when solving. Expanding the perfect square and substituting the final roots provide independent checks.

Audit normalization, balance, factor expansion, and both square-root branches.

Coefficient matching requires an identity that holds for every input. Expanding the perfect square and substituting the final roots provide independent checks.