Roots of a Quadratic Equation

(x^2+2x-8=(x+4)(x-2)) so the roots are (-4) and (2). Roots have opposite signs to the factor constants.

Taking square root of both sides of \(x^2=16\) gives \(x=\pm\sqrt{16}=\pm4\). So both +4 and -4 are roots. Option B is wrong because it omits the negative root (-4); C and D are incorrect since \(\sqrt{16}=4\), not 8, and 0 and 16 do not satisfy the equation. Exam tip: whenever you take square roots in equations, include the \(\pm\) sign to capture both roots.

Let the roots be α and β with product αβ = 12, and given α = 3. Thus β = \\(\frac{12}{3}=4\\). More generally for ax^2+bx+c=0 the product of roots equals c/a. Option A (3) is incorrect because 3×3 = 9 ≠ 12; options C and D (9 and 12) give products 27 and 36 respectively, so they also do not match. Exam tip: use product = (given product) ÷ (given root) to find the other root quickly and check by multiplication.

Let the roots be α and β with α+β = 5, and suppose one root α = 2. Then the other root is β = 5−2 = 3. Option A is wrong because if both roots were 2 the sum would be 4; options C and D do not satisfy the given sum 5. Exam tip: subtract the known root from the total sum and verify by adding the two roots.

If the roots are (r) and (\frac{1}{r}) their product is (1). For reciprocal roots remember the product is (1).

(2x^2-5x+3=(2x-3)(x-1)) so the roots are (\frac{3}{2}) and (1). Be careful with factors that contain coefficients.

A is correct because \(x^2+5x = x(x+5)\), so \(x=0\) is a root. Rule: a quadratic has 0 as a root only if its constant term is 0 (i.e., the polynomial is divisible by x). For contrast, option C factors as \((x-2)(x-3)\) and D as \((x+1)^2\), so their roots are 2, 3 and −1 respectively, not 0. Exam tip: check the constant term quickly — if it is 0, then x=0 is a root.

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).