Introduction to Quadratic Equations

The sum of roots is (2+(-3)=-1). When adding integers with unlike signs, take the sign of the larger magnitude.

The product of roots is (4\cdot6=24). Sum and product are both useful while forming an equation from roots.

For a monic quadratic with root-sum S and product P the equation is \(x^2-Sx+P=0\). With S=5 and P=6 this gives \(x^2-5x+6=0\), so option A is correct. Check distractors: B has product 6 but sum \(-5\), while C and D have product 5, not 6. Exam tip: remember for \(ax^2+bx+c=0\) the sum of roots = \(-b/a)\) and product = \(c/a)\).

The sum of roots is (-\frac{b}{a}=-\frac{-5}{2}=\frac{5}{2}). Pay attention to the sign of (b) in the formula.

For a quadratic \(ax^2+bx+c=0\), the product of roots equals \(c/a\) (Vieta's formula). Here \(a=3\) and \(c=-8\), so the product is \(\dfrac{-8}{3}=-\tfrac{8}{3}\). Option B (\(\tfrac{8}{3}\)) is a sign error; options C and D have incorrect numerical values. Exam tip: apply \(c/a\) directly and pay attention to the sign of \(c\) from the given equation.

For a quadratic \(ax^2+bx+c=0\), the product of the roots equals \(\dfrac{c}{a}\). Here \(a=k\) and \(c=2\), so \(\dfrac{2}{k}=1\) which gives \(k=2\). Option 1 might arise from mistakenly taking \(c\) alone instead of \(c/a\); options -2 and 4 do not satisfy the relation \(2/k=1\). Exam tip: always identify \(a\) (coefficient of \(x^2\)) before applying the product formula \(c/a\).

Putting (x=1) gives (1+1+1=3\neq 0). To check when a value is not a root, use substitution too.

The sum of the roots is 4+4 = 8. For a quadratic \(ax^2+bx+c=0\), the sum of roots equals \(-\frac{b}{a}\). Here \(a=1, b=p\), so \(-p=8\) and hence \(p=-8\). The product check also matches: \(4\times4=16=c/a\). The common wrong choice 8 comes from forgetting the negative sign in the relation. Exam tip: use sum = -b/a and product = c/a and verify both to avoid sign errors.

To simplify the equation, first expand the square of the binomial. The identity \\(x+2)^2 = x^2 + 4x + 4\\) is used because the middle term is twice the product of x and 2. The remaining terms are then combined like ordinary algebraic terms. Care is needed because the expression contains subtraction of both (x+2) and 6.

Substituting the expansion gives x^2 + 4x + 4 - x - 2 - 6 = 0. Combining like terms produces x^2 + 3x - 4 = 0. Therefore option A is correct. The coefficient of x is 3 because 4x - x = 3x, and the constant is -4 because 4 - 2 - 6 = -4. The other options result from expanding or combining the terms incorrectly.

The left side gives (x^2+5x-6), and subtracting (2x) gives (x^2+3x-6=0). First expand and then bring all terms to one side.

Substitute directly: \(4^2-9\cdot4+20=16-36+20=0\). Thus the left-hand side equals 0 and \(x=4\) is a root of the equation. Option B (4) is incorrect because it confuses the variable's value with the value of the polynomial; the polynomial evaluates to 0, not 4. Exam tip: perform arithmetic step by step after substitution to avoid sign errors.

(x^2-16=0) gives (x=\pm4), which are opposite numbers. Pure quadratics often give opposite roots.

For equal roots, (D=0) gives (k^2=100), and for equal negative roots (-\frac{k}{2}<0) is needed. Hence (k=10) is correct.

In (x^2-49=0), the (x^2) term is present and the (x) term is absent. An equation can be quadratic even without an (x) term.

The direct answer is option A, \(x^2-4x+4=0\). Dividing an equation by 3 means dividing every term by 3, not just one term: \(3x^2/3=x^2\), \(-12x/3=-4x\), and \(12/3=4\). The right side remains \(0/3=0\). Therefore the result is \(x^2-4x+4=0\). Option A is correct, and the roots remain unchanged because 3 is non-zero. Option B, \(x^2+4x+4=0\), has the wrong sign for the middle term; −12 divided by 3 is −4, not +4. Option C, \(3x^2-4x+4=0\), fails to divide the first term by 3. Option D, \(x^2-12x+12=0\), divides only the first term and leaves the other coefficients unchanged. A common mistake is to divide selected terms; always apply the operation to every term. Memory cue: divide each coefficient by the same non-zero number.

We need two numbers whose product is 30 and whose sum is 11. \(5\cdot6=30\) and \(5+6=11\), so the quadratic factors as \((x-5)(x-6)\) and the roots are \(x=5\) and \(x=6\). The closest distractor, \(3\) and \(10\), also multiply to 30 but sum to 13, so it does not match the middle coefficient. Exam tip: For \(ax^2+bx+c\), look for factor pairs of \(c\) whose sum equals \(b\), remembering to account for signs of the coefficients.

Use Vieta's relations. For \(x^2+ax+10=0\), the sum of roots equals \(-a\) and the product equals \(10\). The given roots sum to \(-2)+(-5)=-7\, so \(-a=-7\) and hence \(a=7\). Check the product: \((-2)(-5)=10\) matches the constant term, confirming consistency. The closest distractor \(-7\) stems from reversing the sign convention (thinking sum equals \(a\) instead of \(-a\)). Exam tip: For \(x^2+bx+c=0\) remember sum = \(-b\) and product = \(c\) to avoid sign mistakes.

For a square with side \(x\), area = \(x^2\). The statement "7 more than 6 times its side" means area = \(6x+7\), so \(x^2=6x+7\). Rearranging gives \(x^2-6x-7=0\), which matches option A. The closest distractor C (\(x^2-7x-6=0\)) would correspond to area = \(7x+6\), which swaps the coefficients and is not what the question states. Option B has incorrect signs and option D has a different leading coefficient, so both are incorrect. Exam tip: Translate phrases like "a more than b times" as \(b x + a\) to avoid sign or order mistakes.

Use the identity \((a-b)^2=a^2-2ab+b^2\). With \(a=x, b=2\) we get \((x-2)^2=x^2-2\cdot x\cdot2+2^2=x^2-4x+4\), so the left-hand side is the perfect square \((x-2)^2\). The closest distractor, \((x+2)^2\), expands to \(x^2+4x+4\) and therefore has the wrong sign on the middle term. The other options also do not match both the middle and constant terms. Exam tip: match the middle term as \(\pm2ab\) to decide the sign in \((a\pm b)^2\).

Here ((x+3)^2=x^2+6x+9), and bringing all terms to one side gives (x^2+4x-6=0). Use the square identity and transposition carefully.

Expanding \((3x+2)(x-5)\) gives \(3x^2-15x+2x-10\). Combining like terms \(-15x+2x=-13x\) yields the standard form \(3x^2-13x-10=0\). Option C shows an intermediate grouping without combining the like x-terms; options B and D have incorrect signs or constants. Exam tip: always expand first, then combine like terms and write coefficients in descending powers of x.

For a quadratic equation, the coefficient of (x^2) must not be (0). Thus (2m-3\neq 0), so (m\neq \frac{3}{2}).

Using the standard form ax^2 + bx + c = 0, the coefficients are a = -4, b = 6, c = -9. Compute a - b + c = (-4) - 6 + (-9) = -19. The choice -7 commonly results from mistakenly calculating a + b + c instead of a - b + c. Exam tip: write down a, b, c with their signs before substituting to avoid sign errors.

A quadratic equation in standard form is ax² + bx + c = 0, and the question specifically requires the coefficient of x² to be positive. Start with 8x − x² = 15. Move 15 to the left: 8x − x² − 15 = 0, or equivalently −x² + 8x − 15 = 0. Since the x² coefficient is negative, multiply the entire equation by −1. This changes every sign and gives x² − 8x + 15 = 0. Therefore option A is correct. It is essential to change all terms, not just the x² term. Options B, C and D contain one or more incorrect signs and are not equivalent to the original equation.

Substitute \(x=-1\): \((-1)^2+5(-1)+4=1-5+4=0\). Therefore \(x=-1\) is a root. Alternatively, factorization \(x^2+5x+4=(x+1)(x+4)\) shows the roots are \(x=-1\) and \(x=-4\). Option C is incorrect because at \(x=4\) the left side is \(16+20+4\neq0\). Exam tip: verify roots by substitution or by factoring the quadratic.