Roots of a Quadratic Equation

By the factor theorem, if \(x-a\) is a factor then substituting \(x=a\) into the polynomial gives value 0. Hence \(a\) is a root (zero) of the polynomial. Option A (coefficient) is incorrect because a coefficient is the multiplier of a term and does not indicate a value that makes the polynomial zero. Option D (constant term) is a different concept — the term independent of \(x\). Exam tip: plug in \(x=a\); if the polynomial evaluates to 0, then \(x-a\) is a factor.

Substitute the root into the equation: with \(x=3\) we get \(3^2+3p-6=0\), so \(9+3p-6=0\) ⇒ \(3+3p=0\) ⇒ \(p=-1\). A common wrong choice (1) comes from an arithmetic or sign mistake when simplifying; check signs carefully. Exam tip: always substitute the given root directly and simplify step by step to avoid sign errors.

From (5x^2=0) we get (x^2=0) so the root (0) occurs twice. For a zero root also check the factor.

Factor the quadratic: \(x^2-2x = x(x-2)\). Setting each factor to zero gives the roots \(x=0\) and \(x=2\). Since only \(2\) appears among the options, it is the correct choice. Check a common distractor: for \(x=3\), \(3^2-2\cdot3=9-6=3\neq0\), so 3 is not a root. Exam tip: factorise first and then set each factor equal to zero to find roots quickly.

For a quadratic with roots α and β the monic form is x^2-(α+β)x+αβ=0. Here α=4 and β=−1 so α+β=3 and αβ=−4. Substituting gives x^2-3x-4=0, which factors as (x-4)(x+1)=0. The closest distractor C, x^2-5x+4=0, has roots 1 and 4 — it shares root 4 but not −1, so it is incorrect. Exam tip: compute sum and product of the given roots and form x^2-(sum)x+(product)=0.

The sum of roots is (-\frac{b}{a}) so here it is (-7). This can be found quickly without factorising.

For a quadratic \(ax^2+bx+c=0\), the product of roots equals \(\frac{c}{a}\). Here \(a=2\) and \(c=-2\), so product = \(\frac{-2}{2}=-1\). Option B (1) is wrong due to incorrect sign; option C (-2) wrongly uses \(c\) instead of \(c/a\). Exam tip: Always identify \(a\) and \(c\) first and be careful with signs.

Substitute \(x=0\) into \(ax^2+bx+c=0\): this gives \(c=0\). Therefore \(x=0\) is a root only when the constant term \(c\) is zero. If \(c=0\), the equation reduces to \(ax^2+bx=0\) or \(x(ax+b)=0\), so \(x=0\) is indeed a root. Option B is incorrect because \(a=0\) makes the equation linear (not necessarily giving zero as a root); option C (\(b=0\)) does not ensure zero is a root unless \(c=0\) also; option D (\(a=b\)) is irrelevant to whether zero is a root. Exam tip: always test a suspected root by direct substitution into the equation.

If a number is a root of a polynomial equation, substituting that number into the polynomial must make the result zero. The given root is -2, so replace every x in x^2 + kx + 4 = 0 by -2. Substitution must preserve the signs: (-2)^2 is positive 4, while k(-2) is -2k. This produces an equation involving only k.

The substitution gives 4 - 2k + 4 = 0, or 8 - 2k = 0. Hence -2k = -8 and k = 4. Therefore option B is correct. A common error is to treat (-2)^2 as -4, but squaring a negative number gives a positive result. Another error is to write +2k instead of -2k. The supplied answer correctly handles both signs.

Factor the quadratic: \(x^2+6x+9=(x+3)^2\). Hence \((x+3)^2=0\) gives the repeated root \(x=-3\). The discriminant \(\Delta=b^2-4ac=36-36=0\) confirms a double root. Option A is the sign-mistake counterpart; options C and D do not satisfy the sum and product required by the coefficients. Exam tip: check \(\Delta\) first—if it is zero, the equation has one repeated root.

Factor the left side: \(3x^2-12x=3x(x-4)\). A product is zero when one factor is zero, so \(3x=0\) gives \(x=0\) and \(x-4=0\) gives \(x=4\). Hence the roots are 0 and 4. The closest distractor B (3 and 4) is wrong because substituting \(x=3\) yields \(3(3)^2-12(3)=27-36=-9\), not zero. Exam tip: always factor out the common factor first; if you divide by x, check x=0 separately.

The polynomial factors as \(x^2-10x+25=(x-5)^2\), so the repeated root is \(5\). The closest distractor, \(-5\), would be correct only if the factor were \((x+5)^2\), which it is not. Quick exam tip: check for a perfect square or compute the discriminant \(b^2-4ac\); if it equals zero you have a repeated root.

Factor the equation: \((x-4)(x+4)=0\), so the roots are \(x=4\) and \(x=-4\). Substituting \(x=0\) gives \(0^2-16=-16\neq0\), so 0 is not a root. A common trap is \(\sqrt{16}\), which equals 4 and hence is a root; note that \(\sqrt{16}\) denotes the principal (positive) square root. Exam tip: always check by substitution or factoring to confirm whether a candidate value satisfies the equation.

Zero-product rule: if a product of two factors is zero, at least one factor is zero. So set \(x-2=0\) or \(x+7=0\). From \(x-2=0\) we get \(x=2\); from \(x+7=0\) we get \(x=-7\). Therefore the roots are 2 and −7. The closest distractor (A: 2 and 7) arises from taking the sign of the second factor incorrectly, so it is wrong. Exam tip: set each factor equal to zero separately and check signs of constants carefully.

Compute the sum directly: 1 + (−4) = −3. For a quadratic ax^2+bx+c=0 the sum of roots α+β = −b/a, which agrees with the direct addition here. The closest wrong option, −5, would arise if a student mistakenly treated 1 as −1 and computed (−1)+(−4)=−5. Exam tip: always write the signs explicitly and perform the addition carefully when a negative number is involved.

The standard form is (x^2-(\text{sum})x+\text{product}=0) so (x^2-6x+8=0). The sign of the sum term changes.

If a number is a root of an equation, substituting that number into the equation must make the result zero. Here the given root is \(x=1\). Substitute it into \(2x^2-3x+m=0\): \(2(1)^2-3(1)+m=0\), so \(2-3+m=0\). This simplifies to \(m-1=0\), giving \(m=1\), which is option B.

The important point is that the root condition is checked by direct substitution. No quadratic formula or factorisation is needed. Testing the result, the equation becomes \(2x^2-3x+1=0\), and at \(x=1\) its value is \(2-3+1=0\). Thus the supplied answer is correct and the calculation is complete.

Factorising gives \(x^2-3x+2=(x-1)(x-2)\), so the roots are \(x=1\) and \(x=2\). The closest distractor D contains 2 (which is correct) but also 3 (which is not), so D is wrong. Also the discriminant is \(b^2-4ac=9-8=1\), a positive perfect square, confirming two rational real roots. Exam tip: use Vieta — sum of roots =3 and product =2 to check your answer quickly.

For any polynomial, a simple root \(r\) corresponds to the linear factor \(x-r\). So if the root is \(2\), the guaranteed factor is \(x-2\). Analysis of distractors: \(x+2\) would correspond to root \(-2\); \(2x+1\) corresponds to root \(-\tfrac{1}{2}\); and \(x^2+2\) is a quadratic expression and cannot be the linear factor associated with root \(2\). Exam tip: substitute \(x=2\) into the polynomial — it must give zero if 2 is truly a root.

Since \(x^2\ge 0\), we have \(x^2+4\ge 4>0\), so \(x^2+4\) can never be zero and there are no real roots. Using the quadratic discriminant: \(\Delta=b^2-4ac=0^2-4\cdot1\cdot4=-16<0\), which confirms no real solutions. Common distractor: students may think a quadratic always has two roots — that requires \(\Delta>0\). Exam tip: check \(\Delta\) or the minimum value of \(x^2\) to decide real roots quickly.