Roots of a Quadratic Equation

Dividing \(7x^2=0\) by 7 gives \(x^2=0\). This is \((x-0)^2\), so \(x=0\) is a repeated root (multiplicity 2). Using coefficients \(a=7,b=0,c=0\), the discriminant is \(b^2-4ac=0^2-4\cdot7\cdot0=0\), confirming a double root. The closest distractor C is wrong because substituting \(x=7\) gives \(7\cdot7^2=343\ne0\). Exam tip: check the discriminant or factor out common terms to quickly identify repeated roots.

This is a difference of squares since \(x^2-49=(x-7)(x+7)\). Setting each factor to zero gives roots: from \(x-7=0\) we get \(x=7\), and from \(x+7=0\) we get \(x=-7\). Option B is wrong because using 49 or −49 would imply \(x^2=2401\), which does not satisfy the original equation. Exam tip: when taking square roots remember both signs — from \(x^2=a\) we get \(x=\pm\sqrt{a}\).

Factor the expression: \(x^2+4x = x(x+4)\). By the zero-product property, the roots are \(x=0\) and \(x=-4\). Among the given choices \(-4\) is present, so it is correct. For example, substituting \(x=4\) gives \(4^2+4\cdot4=16+16=32\neq0\), so 4 is not a root. Exam tip: always try factoring out the common factor first to apply the zero-product rule quickly.

For a monic quadratic with roots r1 and r2 use \(x^2-(r1+r2)x+r1r2=0\). Here r1=5 and r2=−3 so sum = \(5+(-3)=2\) and product = \(5\times(-3)=-15\). Thus the equation is \(x^2-2x-15=0\) (option A). The closest distractor is option C, \(x^2-8x+15=0\), whose sum is 8 and product 15, giving roots 3 and 5 — it shares root 5 but the other root is different. Exam tip: compute sum and product of given roots and form \(x^2-( ext{sum})x+( ext{product})=0\).

For a quadratic \(ax^2+bx+c=0\), the sum of the roots equals \(-\frac{b}{a}\). Here \(a=1\) and \(b=11\), so the sum is \(-\frac{11}{1}=-11\). Option A (11) is incorrect because the sign must be negated; options C and D correspond to the constant term 30 (or its negative) and are not the sum. Exam tip: quickly identify a and b and apply \(-\frac{b}{a}\).

For a quadratic \(ax^2+bx+c=0\) the product of roots is \(\alpha\beta=\dfrac{c}{a}\). Here \(a=5\) and \(c=-8\), so the product is \(-\dfrac{8}{5}\). Option B has the wrong sign; options C and D are incorrect numerical values. Exam tip: identify \(a\) and \(c\) first and directly compute \(c/a\) to avoid sign mistakes.

If \(x=0\) is a root, substitute \(x=0\) into \(2x^2+bx+c\). That gives \(2(0)^2+b(0)+c=c\), so \(c=0\). Option C (\(c=b\)) is incorrect because the coefficient \(b\) need not equal the constant term without extra conditions. Options B and D are arbitrary numeric values and are not determined by the given information. Exam tip: For a zero root, check the constant term first—it must be zero.

The governing principle is that every root satisfies the equation when substituted for the variable. Since −1 is a root, put x = −1 into x² + rx − 12 = 0. This gives (−1)² + r(−1) − 12 = 0, so 1 − r − 12 = 0. Simplifying, −r − 11 = 0, and therefore r = −11. Hence option A is correct. The crucial detail is the sign of rx: when x = −1, the term r x becomes −r, not +r. Option B is produced by mishandling that sign. The values 12 and −12 are the constant term or its opposite, not the unknown coefficient r. Direct substitution both derives and verifies the answer.

The quadratic is a perfect square: \(x^2+12x+36=(x+6)^2\). Hence both roots are equal and each root is \(-6\). You can also use the discriminant: \(b^2-4ac=12^2-4\cdot1\cdot36=144-144=0\), which confirms a repeated (double) root. Option A (6,6) is a sign error — the correct factor is \(x+6\), not \(x-6\). Exam tip: try factoring first; if unclear compute the discriminant — zero indicates a repeated root.

Factor out the common term: \(2x^2-10x=2x(x-5)\). By the zero-product property set each factor to zero: \(2x=0\) gives \(x=0\), and \(x-5=0\) gives \(x=5\). Hence the roots are 0 and 5. Option B (2 and 5) is incorrect because \(x=2\) does not satisfy the equation; option C has the wrong sign for one root and option D's second root 10 is incorrect. Exam tip: always factor out the greatest common factor first, then apply the zero-product rule and check signs.

Factorizing \(x^2+14x+49\) gives \((x+7)^2\), so the equation \((x+7)^2=0\) has the single repeated root \(-7\). Checking the discriminant: \(b^2-4ac=14^2-4\cdot1\cdot49=0\) confirms one repeated root. Option A (7) would correspond to \((x-7)^2\) and is therefore incorrect; options C and D do not satisfy the given quadratic. Exam tip: either factor into a perfect square or compute the discriminant to quickly detect a repeated root.

Solving \(x^2-64=0\) gives \(x^2=64\), so \(x=\pm8\). Thus 8 and -8 are roots. Also \(\sqrt{64}=8\), so option D represents 8 and is a root. Substituting 0 does not satisfy the equation, so 0 is not a root. Exam tip: for equations of the form \(x^2=a\) remember the solutions are \(x=\pm\sqrt{a}\); \(\sqrt{a}\) by itself denotes the principal (positive) square root.

By the zero-product rule, if a product of two factors is zero then at least one factor must be zero. For (x+1)(x-8)=0 either x+1=0 or x-8=0. Solving gives x=-1 and x=8, so the roots are -1 and 8. The nearest distractor A (1, 8) has the sign of the first root wrong. Exam tip: set each factor to zero to find roots and substitute them back into the equation to verify quickly.

Add the two roots directly: \(2 + (-5) = -3\), so the correct answer is \(-3\). A common mistake is to add absolute values to get \(7\) and then apply a wrong sign, producing \(-7\), which is incorrect. Exam tip: for a quadratic \(ax^2+bx+c=0\) you can quickly check the sum of roots using \(-b/a\).

For a monic quadratic the standard relation is \(x^2-(\text{sum})x+\text{product}=0\). Here sum = −9 and product = 20. Substituting gives \(x^2-(-9)x+20 = x^2+9x+20=0\), so option A is correct. Option B would correspond to sum = +9 (wrong sign). Options C and D have incorrect combinations of sum and product. Exam tip: always use \(x^2-(\text{sum})x+\text{product}=0\) and be careful with the sign of the given sum of roots.

A root must satisfy the equation. Substitute x = 2 into \(3x^2 - 8x + n = 0\): \(3(2)^2 - 8(2) + n = 0\) ⇒ \(12 - 16 + n = 0\) ⇒ \(n = 4\). Option C (\(-4\)) is wrong because it gives \(12 - 16 - 4 = -8 \neq 0\). Exam tip: always substitute the root directly and perform arithmetic carefully to avoid sign errors.

Factorise: \(x^2-5x+6=(x-2)(x-3)\). The roots are values of x that make a factor zero: \(x=2\) and \(x=3\). The closest distractor \((-2,-3)\) is wrong because the signs are reversed — their sum is \(-5\), not the required \(+5\). Exam tip: For quadratics with integer coefficients, look for factor pairs of the constant term whose sum equals the coefficient of x (with sign).

By the factor theorem, if \(r\) is a root of a polynomial then \(x-r\) is a factor. Here \(r=-3\), so the guaranteed factor is \(x-(-3)=x+3\). Option A (\(x-3\)) corresponds to root 3, not -3. Option C (\(3x-1\)) has root 1/3, unrelated to -3. Option D (\(x^2-9\)) equals \((x-3)(x+3)\); it would be a factor only if the other root were 3, so it is not guaranteed. Exam tip: for any root \(r\) use the linear factor \(x-r\); any nonzero constant multiple of that linear factor also vanishes at \(r\), but the canonical factor is \(x-r\).

For a quadratic \(ax^2+bx+c=0\), the number of real roots depends on the discriminant \(D=b^2-4ac\). Here \(a=1,\;b=0,\;c=16\), so \(D=0^2-4\times1\times16=-64<0\). A negative discriminant means no real roots. Alternatively, since for any real x, \(x^2\ge0\), \(x^2+16\) cannot be zero. Exam tip: check the discriminant first — \(D>0\) gives 2 roots, \(D=0\) gives 1, \(D<0\) gives 0 real roots. The option “cannot be determined” is wrong because all coefficients are given.

Factor the quadratic: \(x^2 - 3x - 10 = (x-5)(x+2)\). Hence the solutions are \(x=5\) and \(x=-2\). Check via sum and product: sum = \(5+(-2)=3\) and product = \(5\times(-2)=-10\), matching the equation's coefficients. Closest distractor B (2 and -5) gives the correct product but wrong sum (\(-3\)), so it is incorrect. Exam tip: either factor quickly or use sum/product (Vieta) to verify roots under time pressure.

Solve by taking square root of both sides: \(x=\pm\sqrt{49}=\pm7\). Thus the roots are 7 and -7. Option B is wrong because it ignores the negative root; option C is wrong because 14 arises from a mistaken doubling (e.g. thinking 7+7) rather than taking the square root; option D is unrelated to solving \(x^2=49\). Exam tip: for equations of form \(x^2=a\) with \(a>0\), always include both signs \(\pm\) when writing the roots.

If the roots are α and β, their product is αβ = 35 and one root α = 5, so the other root is β = \(\dfrac{35}{5}=7\). Option A is wrong because if the other root were 5 the product would be \(5\times5=25\). Options C and D are incorrect because they do not equal the result of dividing the product by the given root. Exam tip: For quadratic roots use relations αβ = c/a — to find one root divide the product by the known root.

Let the roots be α and β with α + β = 8. Given one root is 3, the other is β = 8 − 3 = 5. Option A is incorrect because it repeats the given root; C is wrong because 8 is the sum, not an individual root; D is incorrect because it adds instead of subtracting. Exam tip: If the sum of roots and one root are given, find the other root by subtracting the given root from the sum.

The sum of the roots is \(r+(-r)=r-r=0\). Therefore, the correct answer is 0. Option B incorrectly treats both roots as positive and adds them as \(r+r=2r\). In exams, check the signs of the roots before adding them.

Factorization method: compute \(a\cdot c=3\cdot3=9\) and note the middle coefficient is -10, so find two numbers with product 9 and sum -10 — they are -9 and -1. Rewrite: \(3x^2-10x+3=3x^2-9x-x+3=3x(x-3)-1(x-3)=(3x-1)(x-3)\). Hence roots are \(x=\frac{1}{3}\) and \(x=3\). Option B (1, 3) is wrong because substituting \(x=1\) gives \(3-10+3=-4\), not zero. Exam tip: use the product-sum (ac method) and factor by grouping, and always verify roots by substitution or the quadratic formula if unsure.