Introduction to Quadratic Equations

Here ((2x+5)(3x-4)=6x^2+7x-20) and (4(x-1)=4x-4). Bringing all terms to one side gives (6x^2+3x-16=0).

For the equation to be quadratic, (k^2-25\neq0) is required. So both (k\neq5) and (k\neq-5) are necessary.

Expand the squares: \((x+2)^2=x^2+4x+4\) and \(2(x-3)^2=2x^2-12x+18\). Thus, \(3x^2-8x+22=35\), and moving 35 to the left gives \(3x^2-8x-13=0\). Therefore, option A is correct. Exam tip: when converting to standard form, keep all terms on one side and make the other side zero; the constant becomes \(22-35=-13\).

Multiplying the whole equation by (10) gives (4x^2-2+5x+15=70). Therefore the standard form is (4x^2+5x-57=0).

A root must satisfy the equation. Substituting \(x=-3\) gives \(4(-3)^2+p(-3)-9=0\), so \(36-3p-9=0\). Hence, \(27-3p=0\) and \(p=9\). Therefore, option A is correct. Exam tip: Substitute the given root directly into the equation and solve for the unknown parameter.

A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-11) gives (x^2+11x+30=0).

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=-2m\), and \(c=36\), so \(D=b^2-4ac=(-2m)^2-4(1)(36)=4m^2-144\). Setting \(D=0\) gives \(4m^2-144=0\), hence \(m^2=36\) and \(m=\pm6\). Choosing only \(m=6\) or only \(m=-6\) omits one valid possibility. Exam tip: whenever the roots are equal, set the discriminant equal to zero.

A quadratic equation has real roots when its discriminant satisfies \(D\geq 0\). Here, \(a=1\), \(b=2p\), and \(c=25\), so \(D=(2p)^2-4(1)(25)=4p^2-100\). Thus, \(4p^2-100\geq 0\), which gives \(p^2\geq 25\), and hence \(p\leq -5\) or \(p\geq 5\). Option B reverses the required inequality and also excludes the double real roots occurring at \(p=\pm5\). Exam tip: For real roots, use \(D\geq0\), not only \(D>0\).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here it is (\frac{\frac{13}{4}}{\frac{9}{4}}=\frac{13}{9}).

For the quadratic equation \(x^2-sx+p=0\), the sum of the roots is \(s\), and their product is \(p\). Thus, \(s=5+6=11\) and \(p=5\times6=30\), so \(s+p=11+30=41\). Exam tip: Compare the equation with \(x^2-(\text{sum of roots})x+(\text{product of roots})\) to identify the parameters quickly.

For a quadratic equation \(ax^2+bx+c=0\), the roots are real and distinct only when the discriminant \(D=b^2-4ac\) is greater than zero. Here, \(a=1\), \(b=-12\), and \(c=k\), so \(D=(-12)^2-4(1)(k)=144-4k>0\). Therefore, \(k<36\). When \(k=36\), the roots are equal, so that option is not correct. Exam tip: For “real and distinct” roots, always apply \(D>0\).

If roots \(\alpha,\beta\) are reciprocals, then \(\alpha\beta=1\). For \(ax^2+bx+c=0\), their product is \(c/a\). In option A, \(c/a=1/1=1\); in B it is \(1/2\). Exam tip: check \(c/a\) directly.

\(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\). Here \(\left(-\frac{6}{5}\right)^2-2\cdot\frac{1}{5}=\frac{26}{25}\).

By Vieta’s formulas, (\alpha+\beta=9) and (\alpha\beta=18). Therefore, ((\alpha-3)(\beta-3)=\alpha\beta-3(\alpha+\beta)+9=18-3(9)+9=0). Hence, option A is correct. Exam tip: expand the product carefully; subtracting 3 from each root introduces the middle term (-3(\alpha+\beta)), which is often missed.

In option B, bringing all terms to one side gives \(x^2+4x-3x-7=0\), or \(x^2+x-7=0\). Its highest power is 2, so it is quadratic. Option A is linear. Exam tip: first rewrite the equation as \(ax^2+bx+c=0\).

Here, \(a=1\), \(b=-8\), and \(c=17\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-8)^2-4(1)(17)=64-68=-4\). Since \(\Delta<0\), the equation has no real roots, so option A is correct. Option B would be correct only if \(\Delta=0\). Exam tip: \(\Delta<0\) means no real roots, \(\Delta=0\) means equal real roots, and \(\Delta>0\) means distinct real roots.

For a quadratic equation, the product of the roots equals the constant term divided by the coefficient of \(x^2\). Here, the product of the roots is \(5m\). Since one root is \(5\), the other root is \(\frac{5m}{5}=m\). Exam tip: divide the product of the roots by the known root to obtain the other root.

Factoring gives \((3x-8)(x-2)=0\), so the roots are \(x=\frac{8}{3}\) and \(x=2\). Therefore, their absolute difference is \(\left|\frac{8}{3}-2\right|=\frac{2}{3}\). Exam tip: when the difference between roots is asked, use the larger root minus the smaller root unless an order is explicitly specified.

The coefficient of \(x^2\) must be non-zero, so \(a\ne0\) is essential. If \(a=0\), the equation becomes linear at most. Exam tip: identify the equation type by checking the highest power of the variable.

The roots are \\(-8\\) and \\(-8\\), so their sum is \\(-16\\). For \\(x^2+bx+64=0\\), the sum of the roots is \\(-b\\). Therefore, \\(-b=-16\\), giving \\(b=16\\). Option B is the root sum, not the coefficient \\(b\\). Exam tip: In \\(x^2+px+q=0\\), the sum of roots is \\(-p\\) and their product is \\(q\\).

In the first option, the sum is (-\frac{b}{a}=6) and the product is (\frac{c}{a}=-16). So the sum is positive and the product is negative.

Area of a rectangle = length × breadth, so \((x+8)(x-5)=104\). On expanding, we get \(x^2+3x-40=104\). Bringing all terms to one side gives \(x^2+3x-144=0\), so option A is correct. Option B results from failing to transfer the \(-40\) term correctly while simplifying. Exam tip: For area-based word problems, first write length × breadth = given area, then rearrange the equation into standard quadratic form.

If the sum of roots is (15) and product is (54), the equation is (x^2-15x+54=0). Remember the monic form formula.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the value is (\frac{11}{30}).

Expanding gives \\(x+a\\)^2=x^2+2ax+a^2"). Comparing the coefficients of \\(x\\), we get \\(2a=18\\), so \\(a=9\\). This also satisfies the constant-term condition \\(a^2=81\\). Although \\(-9\\) gives the same constant term, it makes the coefficient of \\(x\\) equal to \\(-18\\), so it is incorrect. Exam tip: For an identity, compare coefficients of like powers on both sides.