Roots of a Quadratic Equation

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.

For a repeated root r the monic quadratic is \((x-r)^2=0\). With r=−4 we get \((x+4)^2=0\), which expands to \((x^2+8x+16=0)\). Option B would correspond to repeated root +4 (sign of the linear term is opposite), so it is incorrect. Options C and D have different coefficients and do not give root −4. Exam tip: to form a monic quadratic from roots \(\alpha,\beta\) use \(x^2-(\alpha+\beta)x+\alpha\beta=0\).

Putting (x=\frac{1}{2}) gives (1+\frac{p}{2}-3=0), so (p=4). When substituting a fractional root, solve each term carefully.

When \(p(a)=0\), substituting \(x=a\) satisfies the equation \(p(x)=0\). Therefore, \(a\) is called a root of the equation. Coefficients are numerical factors of terms, a constant is a fixed term, and the degree is the highest exponent. Exam tip: to check whether a value is a root, substitute it into the polynomial and verify that the result is \(0\).