Introduction to Quadratic Equations

The sum of roots is (5+p), and in the equation the sum is (-p). Thus (5+p=-p), giving (p=-\frac{5}{2}).

By Vieta’s formulas, \(\alpha+\beta=16\) and \(\alpha\beta=63\). Therefore, \((\alpha-7)(\beta-7)=\alpha\beta-7(\alpha+\beta)+49=63-7(16)+49=0\). Hence, option A is correct. Exam tip: Use the sum and product of the roots directly instead of finding the roots separately.

The area of a triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Hence, \(\frac{1}{2}(x+4)(x+8)=72\), so \((x+4)(x+8)=144\). Expanding gives \(x^2+12x+32=144\), and therefore \(x^2+12x-112=0\). Option B incorrectly subtracts \(72\) instead of \(144\), failing to account for the factor \(\frac{1}{2}\). Exam tip: In triangle-area problems, do not forget the factor \(\frac{1}{2}\).

By Vieta’s formula, the sum of the roots of \(ax^2+bx+c=0\) is \(-\frac{b}{a}\). Here, \(a=1\) and \(b=-15\), so \(\alpha+\beta=-\frac{-15}{1}=15\). Therefore, \((\alpha+\beta)^2=15^2=225\). The value 44 is the product \(\alpha\beta\), not the sum of the roots. Exam tip: Identify the sum and product of roots directly using Vieta’s formulas.

In option C, expanding gives \(x^2+2x+1=x^2+4x\). The \(x^2\) terms cancel, leaving \(2x-1=0\), which is linear. Exam tip: simplify fully and then check the highest power of x.

The equation can be rewritten as \((x-a)^2-36=0\). Thus, \((x-a)^2=36\), giving the roots \(x=a+6\) and \(x=a-6\). Their difference is \((a+6)-(a-6)=12\). The expression \(a+6\) represents only one root, not the difference. Exam tip: First rewrite such equations in perfect-square form.

The original equation requires x ≠ 1 and x ≠ −2 because these values make a denominator zero. Multiply both sides by 2(x − 1)(x + 2), which is valid under those restrictions. This gives 2(x + 2)² + 2(x − 1)² = 5(x − 1)(x + 2). Expanding the left side: 2(x² + 4x + 4) + 2(x² − 2x + 1) = 4x² + 4x + 10. Expanding the right side gives 5(x² + x − 2) = 5x² + 5x − 10. Moving the left side to the right yields 0 = x² + x − 20, or x² + x − 20 = 0. Thus option A is correct; the other options have an incorrect sign, constant, or leading coefficient.

We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha+\beta=\frac{7}{2}) and (\alpha\beta=\frac{3}{2}), so the value is (\frac{37}{6}).

The sum is (2k+3) and the product is (k^2+3k). Direct checking with the options shows (k=3) gives the required equation pattern.

The sum of roots is (-p), so ((p+1)+(q-1)=p+q=-p). Using (p+q=5), the option consistent with the conditions is (6).

Here ((4x-3)(x+5)=4x^2+17x-15) and (2(3x-1)=6x-2). Bringing all terms to one side gives (4x^2+11x-13=0).

For the equation to be quadratic, the coefficient of (x^2) must not be (0). Here (t^2-64\neq0), so (t\neq\pm8).

Expand the squares first: \(3(x+1)^2=3(x^2+2x+1)=3x^2+6x+3\) and \(2(x-4)^2=2(x^2-8x+16)=2x^2-16x+32\). Adding gives \(5x^2-10x+35=74\). Bring 74 to the left and simplify: \(5x^2-10x+35-74=0\), so \(5x^2-10x-39=0\). Option B is the closest distractor because it has the wrong sign on the linear term (+10x instead of -10x). Exam tip: expand carefully, combine like terms, then move the constant from the RHS to the LHS and re-check the arithmetic for the constant term (here \(3+32-74=-39\)).

Multiplying the whole equation by (20) gives (16x^2-12+5x-10=60). Therefore the standard form is (16x^2+5x-82=0).

Substitute \(x=-6\) into \(2x^2+px-18=0\): \(2(-6)^2 + p(-6) -18 = 0\). This gives \(72 - 6p - 18 = 0\) → \(54 - 6p = 0\), so \(p = 54/6 = 9\). The choice \(-9\) typically arises from a sign error (treating \(p(-6)\) as \(+6p\)). Exam tip: always substitute the root and simplify step-by-step, checking signs carefully.

A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-15) gives (x^2+15x+56=0).

For equal roots the discriminant must be zero: \(D=0\). In the given equation \(a=1,\ b=-2m,\ c=64\). So \(D=b^2-4ac=(-2m)^2-4\cdot1\cdot64=4m^2-256\). Setting \(D=0\) gives \(4m^2-256=0\Rightarrow m^2=64\Rightarrow m=\pm8\). Thus \(m=\pm8\) is correct. The option \(m=\pm16\) is incorrect because it yields a nonzero discriminant (roots are not equal). Exam tip: always compute \(b^2-4ac\) first for equal/distinct roots and solve the resulting equation for the parameter.

For real roots the discriminant must satisfy \(D\ge0\). Here \(D=(2p)^2-4\cdot1\cdot49=4p^2-196=4(p^2-49)\). Requiring \(D\ge0\) gives \(p^2-49\ge0\) or \(p^2\ge49\), hence \(p\le-7\) or \(p\ge7\). Option B is incorrect because for \(-7<p<7\) we have \(p^2<49\) so \(D<0\); option C is just a special value and not the general condition; option D is too broad and includes values that do not give real roots. Exam tip: For parameter problems, compute \(D=b^2-4ac\) first and set \(D\ge0\) to find the range of the parameter.

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here it is (\frac{\frac{19}{6}}{\frac{10}{6}}=\frac{19}{10}).

By Vieta’s relations for \(x^2+bx+c=0\), the sum of roots = \(-b\) and the product = \(c\). For \(x^2 - s x + p = 0\), the sum of the roots equals \(s\) and the product equals \(p\). With roots 8 and 9 we get \(s=8+9=17\) and \(p=8\times9=72\). Hence \(s+p=17+72=89\). Note that 17 is only the sum (s) and 72 is only the product (p), so they are incorrect as s+p. Exam tip: apply Vieta’s formulas directly instead of recomputing coefficients each time.