Roots of a Quadratic Equation

With both roots (7) we get ((x-7)^2=0) which is (x^2-14x+49=0). Form a perfect square from repeated roots.

Putting (x=3) gives (18-21+k=0), so (k=3). In such questions substitute the given root directly into the equation.

If a number \(\alpha\) is a root, then substituting it gives \(p(\alpha)=0\). Thus the value of the polynomial must become zero. Options A and C (1 and -1) could be roots for some specific polynomials but are not the general definition. Option D (\(p(x)\)) is the polynomial expression itself, not the numeric value it attains. Exam tip: always verify a claimed root by substitution — compute \(p(\text{the value})\) and check if it equals 0.

Test by substitution: put \(x=1\) into the left side: \(1^2-3\cdot1+2=1-3+2=0\). Since the expression equals zero, \(x=1\) is a root. The closest distractor (Option C) is wrong because \(x=2\) is also a root, so it is not "only at \(x=2\)". Exam tip: verify a claimed root by direct substitution into the quadratic.

Substituting \(x=-2\) gives \((-2)^2+3(-2)+2 = 4-6+2 = 0\), so \(x=-2\) is indeed a root. Factoring yields \(x^2+3x+2=(x+1)(x+2)\), so the real roots are \(x=-1\) and \(x=-2\). Thus 'Only at \(x=2\)' is incorrect because \(+2\) is not a root, and 'No real root' is also incorrect. Exam tip: either substitute carefully (watch signs) or factor the quadratic to find/check roots quickly.

Write the expression as a difference of squares: \(x^2-25=(x-5)(x+5)\). Setting each factor to zero gives \(x-5=0\Rightarrow x=5\) and \(x+5=0\Rightarrow x=-5\). Thus the roots are 5 and -5. The closest distractor 25 and -25 is wrong because from \(x^2=25\) the solutions are \(x=\pm\sqrt{25}=\pm5\), not \(\pm25\). Exam tip: for equations of the form \(x^2=a\) always take square root to get \(x=\pm\sqrt{a}\).

Factorise the quadratic: \(x^2-7x+12=(x-3)(x-4)\). Setting each factor to zero gives \(x=3\) or \(x=4\), so the roots are 3 and 4. Option B (−3 and −4) also gives product 12 but their sum is −7, not +7; for \(ax^2+bx+c=0\) the sum of roots is \(-b/a\) and product is \(c/a\). Exam tip: find two numbers whose product is 12 and whose sum is 7 — those are the roots.

Factor the quadratic: \(x^2+9x+20=(x+4)(x+5)\). Setting each factor to zero gives the roots \(x=-4\) and \(x=-5\). Option B (4, 5) has the wrong signs — the factors are \(x+4\) and \(x+5\), so the variable equals the negatives of 4 and 5. Exam tip: look for two numbers with product 20 and sum 9, then use signs of the factors to get the correct roots and verify by substitution if unsure.

If a quadratic has roots r_1 and r_2 its monic form is \(k(x-r_1)(x-r_2)=0\). With roots 0 and -6 we get \(x(x+6)=0\), which expands to \(x^2+6x=0\). Option B corresponds to \(x(x-6)=0\) so its roots are 0 and 6, not -6. Option C, \(x^2+36=0\), has complex roots ±6i. Option D, \(x^2-36=0\), has roots ±6 (not 0 and -6). Exam tip: verify quickly by substituting x=0 and x=-6 into the candidate equation — both should satisfy it for a correct choice.

Substitute the given root into the equation: with \(x=2\) we get \(4+2q-10=0\), i.e. \(2q-6=0\). Hence \(q=3\). Options A, C and D are incorrect because they do not satisfy the resulting equation (for example, \(q=2\) gives \(2q-6\neq0\)). Exam tip: always substitute the known root into the quadratic and solve the resulting linear equation for the parameter.

For a quadratic \(ax^2+bx+c=0\) the sum of roots equals \(-\frac{b}{a}\). Here \(a=3\) and \(b=-5\), so the sum is \(-\frac{-5}{3}=\frac{5}{3}\). The nearest distractor \(-\frac{5}{3}\) is a sign error; \(\frac{2}{3}\) is the product of the roots \(c/a\), not the sum. Exam tip: identify \(a,b,c\) first and then use \(-b/a\).

For a quadratic \(ax^2+bx+c=0\), the product of roots equals \(\frac{c}{a}\). Here \(a=4\) and \(c=-3\), so the product is \(\frac{-3}{4}= -\frac{3}{4}\). The closest distractor A (\(\frac{3}{4}\)) is a sign error; options C and D are incorrect because they come from using the wrong coefficient. Exam tip: remember product = \(c/a\) and sum = \(-b/a\) to avoid sign mistakes.

When (D>0), two distinct real roots are obtained. To know the nature of roots, check (D=b^2-4ac).

Use the formula \(D=b^2-4ac\). Here \(a=1,\; b=-2,\; c=5\), so \(D=(-2)^2-4\cdot1\cdot5=4-20=-16\). Thus the correct discriminant is -16; a negative D means no real roots (the roots are complex conjugates). The closest distractor 16 is wrong because it comes from treating b as +2 instead of -2. Exam tip: always substitute b with its sign (the coefficient as written) to avoid sign errors.

The sign of the discriminant D determines the nature of the roots. When \(D<0\) the quantity \(\sqrt{D}\) is not real, so the roots are complex conjugates and there are no real roots. If \(D>0\) there are two distinct real roots, and if \(D=0\) there is one repeated real root; thus those options are incorrect. Exam tip: compute \(b^2-4ac\) quickly — if it is negative you can immediately conclude there are no real roots.

For any real \\((x)\\), \\(x^2\\ge0\\), so \\(x^2+9\\ge9>0\\) and cannot be zero. The discriminant \\(b^2-4ac = 0^2 - 4\cdot1\cdot9 = -36 < 0\\) therefore confirms there are no real roots; the equation has complex roots \\(\pm 3i\\). Closest distractor (option D) is wrong because \\(3^2+9=18\\), not zero. Exam tip: check the discriminant \\(b^2-4ac\\) to decide existence of real roots quickly.

Factor the quadratic: \(x^2-5x = x(x-5)\). If a product is zero, at least one factor is zero, so \(x=0\) or \(x-5=0\) giving \(x=5\). Thus the roots are 0 and 5. The closest distractor is D (0 and -5) — it has 0 correct but the sign of the other root is wrong because the factor is \(x-5\), not \(x+5\). Exam tip: Always factor and set each factor to zero; carefully check signs when solving for roots.

Write the equation as \(x^2-9x=0\) and factor: \(x(x-9)=0\). By the zero-product property either \(x=0\) or \(x-9=0\Rightarrow x=9\). Hence the roots are 0 and 9. A common mistake is dividing both sides by \(x\), which loses the root \(x=0\). Exam tip: bring to standard form and factor, or if you divide by a variable, always check for the lost zero root.

Divide both sides by 3: \(x^2=9\). Taking square roots gives \(x=\pm3\), so the roots are 3 and −3. Option B (9, −9) is incorrect because it confuses the value of \(x^2\) with the values of x (the square roots are ±3, not ±9). Options C and D omit the negative root. Exam tip: first divide by the leading coefficient to get \(x^2=\) form, then apply ±√ to find both roots.

Factor the quadratic: \(x^2-8x+16=(x-4)^2\). Hence \((x-4)^2=0\) gives the repeated root \(x=4\). Alternatively compute the discriminant: \(\Delta=b^2-4ac=(-8)^2-4\cdot1\cdot16=0\), and \(\Delta=0\) indicates a double root. The closest distractor \(-4\) is incorrect because substituting \(-4\) does not satisfy the equation (it gives 64). Exam tip: either factor into a perfect square or check the discriminant; if \(\Delta=0\) the root is repeated.

If a quadratic has roots α and β, it can be written as \((x-α)(x-β)=0\). With roots -2 and 6 we get \((x+2)(x-6)=0\). Expanding gives \(x^2-4x-12=0\), so option A is correct. For comparison, option C expands to \((x-2)(x-6)=x^2-8x+12\) which has roots 2 and 6, so it is wrong. Exam tip: check sum and product of roots quickly — for a monic quadratic \(x^2+bx+c\), sum of roots = -b and product = c; here sum = -2+6=4 (so \(b=-4\)), product = -12.

Use factorisation or Vieta's relations. Factorising gives \(x^2-9x+18=(x-3)(x-6)\), so if one root is 3 the other is 6. By Vieta, the sum of roots is \(\alpha+\beta=9\); with \(\alpha=3\) we get \(\beta=6\). The distractor 3 is incorrect because a repeated root 3 would make the polynomial \((x-3)^2=x^2-6x+9\), which does not match the given polynomial. Exam tip: try quick factorisation first; if not obvious use sum/product of roots.

A positive product means both roots have the same sign. A negative sum means both roots are negative.

If \(x+5\) is a factor, set the factor equal to zero: \(x+5=0\) gives \(x=-5\). Hence the certain root is \(-5\). Option A (5) is incorrect because the sign is reversed; solving \(x+5=0\) does not give +5. Option C (0) would only be correct if the factor were \(x\). Option D (\(\tfrac{1}{5}\)) would arise from a factor \(x-\tfrac{1}{5}\), not \(x+5\). Exam tip: always equate the linear factor to zero to find its root quickly.

Substitute \(x=4\) into the equation: \(16+4m-20=0\), so \(4m-4=0\) and hence \(m=1\). The nearest distractor (\(-1\)) is wrong because with \(m=-1\) the left side becomes \(16-4-20=-8\), not zero. Exam tip: you can also use Vieta — with one root 4 the other is \(-5\) (product = \(-20\)), sum = \(4+(-5)=-1=-m\) giving \(m=1\).