Introduction to Quadratic Equations

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.

Dividing every term by (3) gives (x^2-4x+4=0). Dividing by a common nonzero factor does not change the roots.

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.

Substitute the given root into the equation: \(2(3)^2+q(3)-15=0\) ⇒ \(18+3q-15=0\) ⇒ \(3+3q=0\) ⇒ \(q=-1\). Option A (1) often arises from a sign mistake when moving terms; options C and D come from arithmetic or sign errors. Exam tip: always substitute the root first and simplify step by step, checking basic addition/subtraction carefully.

(4x^2+1=3x) can be written as (4x^2-3x+1=0). In standard form, all terms are on one side and (0) is on the other.

((x+4)^2=x^2+8x+16), and subtracting (5x+10) gives (x^2+3x+6=0). Keep the middle term correct in square identities.

Substituting into (ax^2+bx+c=0) gives (5x^2-11x-6=0). Treat the negative sign as part of the coefficient.

The discriminant is given by \(D=b^2-4ac\). Here \(a=9,\; b=-12,\; c=4\). Thus \(D=(-12)^2-4\cdot9\cdot4=144-144=0\). Option B mistakenly gives only \(b^2\) (144) without subtracting \(4ac\). Option C arises from incorrectly treating \((-12)^2\) as -144; squaring a negative yields a positive. Exam tip: identify \(a,b,c\) first, then compute \(b^2-4ac\) carefully.