Introduction to Quadratic Equations

Here ((5x-2)(2x+3)=10x^2+11x-6) and subtracting (7x) gives (10x^2+4x-6=0). First expand and then bring all terms to one side.

A quadratic equation in x must have a non-zero coefficient of x². Here that coefficient is m² − 9, so the condition is m² − 9 ≠ 0. Factorising gives (m − 3)(m + 3) ≠ 0, which means that m must be different from both 3 and −3. In compact notation, this is m ≠ ±3, so option C is correct. If m = 3 or m = −3, then m² − 9 becomes zero, the x² term disappears, and the equation reduces to 4x − 5 = 0, which is linear. Option A excludes only 3, option B excludes only −3, and option D selects the two values that must be rejected. Therefore, option C is the complete condition.

Expanding gives \((x-2)^2=x^2-4x+4\) and \((2x+1)^2=4x^2+4x+1\). Therefore, \(5x^2+5=25\); moving 25 to the left gives \(5x^2-20=0\). Hence, option D is correct. Exam tip: To obtain standard form, bring all terms to one side and set the resulting expression equal to zero.

Multiplying the whole equation by (6) gives (3x^2-9+2x-2=24). Thus the standard form is (3x^2+2x-35=0).

The value of a root must satisfy the equation. Substituting \\(x=-2\\) gives \\(3(-2)^2+p(-2)+10=0\\), so \\(12-2p+10=0\\), or \\(22-2p=0\\). Hence, \\(p=11\\). In such questions, substitute the given root directly and simplify carefully, especially the sign of the middle term.

If the roots are \(\alpha\) and \(\beta\), the standard form of a monic quadratic equation is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). Here, \(\alpha+\beta=7\) and \(\alpha\beta=10\), so the equation is \(x^2-7x+10=0\). In option B, the sum of the roots would be \(-7\), while option C interchanges the sum and product. Exam tip: in a monic quadratic, the coefficient of \(x\) is the negative of the sum of the roots, and the constant term is their product.

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=-2k\), and \(c=16\), so \(D=b^2-4ac=(-2k)^2-4(1)(16)=4k^2-64\). Thus, \(4k^2-64=0\), giving \(k^2=16\) and hence \(k=\pm4\). Therefore, option A is correct; choosing only \(k=4\) or only \(k=-4\) gives an incomplete answer. Exam tip: For equal roots of a quadratic equation, immediately use \(D=0\).

A quadratic equation has real roots only when its discriminant \(D=b^2-4ac\) is non-negative. Here, \(a=1\), \(b=2k\), and \(c=9\), so \(D=(2k)^2-4(1)(9)=4k^2-36\). Thus \(4k^2-36\geq 0\), which gives \(k^2\geq 9\), and hence \(k\leq -3\) or \(k\geq 3\). In option B, \(D<0\), so the roots are not real. Exam tip: For real roots, always check the condition \(D\geq 0\).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here it is (\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}).

By Vieta’s relations, the sum of the roots equals \(s\), while their product equals \(p\). Thus, \(s=2+3=5\) and \(p=2\times3=6\). Therefore, \(s+p=5+6=11\), so option A is correct. Exam tip: for \(x^2-sx+p\), identify the sum of roots as \(s\) and the product as \(p\).

For a quadratic equation to have real and distinct roots, its discriminant \(D=b^2-4ac\) must be positive. Here, \(a=1\), \(b=-6\), and \(c=k\), so \(D=(-6)^2-4(1)(k)=36-4k\). Thus, \(36-4k>0\), which gives \(k<9\). At \(k=9\), the roots are equal, so option B is not correct. Exam tip: for two distinct real roots, always require \(D>0\).

If the roots are \(\alpha,\beta\), then \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\). Here \(\left(-\frac{7}{2}\right)^2-2\cdot\frac{3}{2}=\frac{37}{4}\).

By Vieta’s formulas, \(\alpha+\beta=8\) and \(\alpha\beta=15\). Therefore, \((\alpha-1)(\beta-1)=\alpha\beta-(\alpha+\beta)+1=15-8+1=8\). Hence, 8 is correct. In such questions, using the sum and product of the roots directly is faster and safer than solving for the roots individually.

The degree of an equation in x is determined by the highest power of x whose coefficient is nonzero. A quadratic equation must therefore have a nonzero coefficient of x². In option A, t = 2 makes the coefficient t − 2 equal to zero. The equation becomes 5x + 1 = 0, whose highest power of x is 1; it is linear and no longer quadratic. Thus option A is correct. In option B, t = 0 changes the x coefficient but leaves the x² coefficient equal to 3. In option C, the coefficient of x² is 1, and in option D it is 2. Those equations remain quadratic. A parameter in a lower-degree term does not alter the degree unless it causes the leading x² coefficient to vanish.

Expanding gives \((x+1)(x+2)=x^2+3x+2\) and \((x+3)(x+4)=x^2+7x+12\). Thus, the left side becomes \(2x^2+10x+14\). Bringing 50 to the left gives \(2x^2+10x+14-50=0\), or \(2x^2+10x-36=0\). Option C is incorrect because dividing the entire equation by 2 would give a constant term of \(-18\), not \(-36\). Exam tip: Write a quadratic equation in standard form as \(ax^2+bx+c=0\).

For a monic quadratic \(x^2+px+q=0\), the sum of the roots is \(-p\), and their product is \(q\). The question says that the roots themselves are \(p\) and \(q\). Substituting these names into the root relations gives \(p+q=-p\) for the sum and \(pq=q\) for the product. These are exactly the two statements in option A.

The symbols \(p\) and \(q\) play two roles here: they are coefficients and, according to the condition, also root values. That is unusual but valid. The relations do not require solving for particular numerical values; they follow directly from Vieta’s formulas. Options B, C, and D alter one or both signs or products, so they do not match the standard coefficient-root relations. Hence A is correct.

Here, \(a=1\), \(b=-4\), and \(c=6\). The discriminant is \(D=b^2-4ac=(-4)^2-4(1)(6)=16-24=-8\), which is less than zero. Therefore, the equation has no real roots; it has a pair of complex conjugate roots. Option B would require \(D=0\). Exam tip: determine the sign of the discriminant first—when \(D<0\), there are no real roots.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, the product of the two roots is \(2m\). Since one root is \(2\), the other root is \(\frac{2m}{2}=m\). As a check, the sum of the roots is \(m+2\), and \(2+m=m+2\), so the result is consistent. Exam tip: When one root is known, use either the product or the sum of roots, whichever gives the quicker calculation.

Factoring the equation as \((3x-4)(x-2)=0\) gives the roots \(x=\frac{4}{3}\) and \(x=2\). Therefore, their positive difference is \(2-\frac{4}{3}=\frac{2}{3}\). In an exam, factoring first and subtracting the smaller root from the larger one is the quickest method.

For the quadratic expression to be a perfect square, \(x^2+6x+c\) must equal \((x+3)^2\). Since \((x+3)^2=x^2+6x+9\), we get \(c=9\), and the equation becomes \((x+3)^2=0\). In the exam, identify the pattern \(x^2+2ax+a^2=(x+a)^2\).

The sum of the two roots is \((-5)+(-5)=-10\). For \(x^2+bx+25=0\), the sum of the roots is \(-b\). Hence, \(-b=-10\), giving \(b=10\). Exam tip: for \(x^2+bx+c=0\), remember that the sum of the roots is \(-b\) and their product is \(c\).

For \(ax^2+bx+c=0\), the sum of roots is \(-\frac ba\), and their product is \(\frac ca\). In option A, the equation is \(x^2-5x+6=0\), so \(a=1\), \(b=-5\), and \(c=6\). The sum is \(-(-5)/1=5\), and the product is \(6/1=6\). Both values are positive.

For comparison, option B has a negative root sum, while options C and D have negative root products because their constant terms are negative. Therefore only option A satisfies both requested conditions. The actual roots are 2 and 3, which also confirms a positive sum and positive product. Thus the supplied answer and explanation are correct.

The area of a rectangle is length × breadth, so ((x+5)(x-2)=84). Expanding the product gives (x^2+3x-10=84). Bringing 84 to the left side results in (x^2+3x-94=0), so option A is correct. Option B incorrectly changes the constant term. Exam tip: expand the binomials carefully and then rearrange the equation into the standard form equal to zero.

If the sum of roots is (11) and product is (30), the equation is (x^2-11x+30=0). Remember the monic form formula.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the value is (\frac{5}{6}).