Roots of a Quadratic Equation

For a quadratic equation \(ax^2+bx+c=0\), here \(a=3\), \(b=-6\), and \(c=3\). Thus, \(D=b^2-4ac=(-6)^2-4(3)(3)=36-36=0\). Therefore, the two roots are real and equal. Exam tip: when \(D=0\), a quadratic equation has two equal real roots.

For the given quadratic equation, \(a=1\), \(b=-4\), and \(c=13\). Its discriminant is \(D=b^2-4ac=(-4)^2-4(1)(13)=16-52=-36\). Since \(D<0\), the equation has no real roots. Equal real roots would require \(D=0\), so option B is incorrect. Exam tip: use the sign of the discriminant to determine the nature of the roots quickly.

For a monic quadratic with root sum \(S\) and product \(P\), the equation is \(x^2-Sx+P=0\). Substituting \(S=5\) and \(P=6\) gives option A. Option B has root sum \(-5\). Exam tip: check the sign of the middle term carefully.

If the sum of the roots is \(S\) and their product is \(P\), the monic quadratic equation is \(x^2-Sx+P=0\). Substituting \(S=9\) and \(P=18\) gives \(x^2-9x+18=0\). Option B has the wrong sign for the sum, while options C and D interchange the sum and product. Exam tip: use a negative coefficient for the sum of roots and a positive constant equal to their product.

For a quadratic equation \(ax^2+bx+c=0\), the product of the roots is \(\frac{c}{a}\). Here, \(a=1\) and \(c=54\), so the product of the two roots is \(54\). If one root is \(6\), the other root is \(\frac{54}{6}=9\). Option 6 merely repeats the given root, while 15 is the sum of the roots, not their product. Exam tip: use the product-of-roots relation and divide by the known root.

The sum of new roots is (3\alpha+3\beta=3(\alpha+\beta)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.

The direct answer is option A: 25. The required identity is α² + β² = (α + β)² - 2αβ. It is obtained by expanding the square of the sum: (α + β)² = α² + 2αβ + β². Now use the given values. The sum is 7, so (α + β)² = 7² = 49. The product is 12, so 2αβ = 2 × 12 = 24. Subtract: α² + β² = 49 - 24 = 25. Hence option A is correct. Option B, 37, would result from an incorrect subtraction, such as 49 - 12, where the factor 2 is forgotten. Option C, 49, is just the square of the sum and does not equal the sum of the separate squares because the extra term 2αβ is present. Option D, 84, is not produced by the identity and likely comes from an incorrect multiplication or addition. There is no need to solve for the individual roots. Memory cue: write the pattern (a² + b² = (a+b)² - 2ab) before inserting numbers, so the factor 2 is not missed.

Here (\alpha+\beta=-5) and (\alpha\beta=-14). Therefore (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}).

The sum is (-\frac{b}{a}=\frac{14}{3}) and the product is (\frac{c}{a}=\frac{8}{3}). Identify (a), (b), and (c) first.

The roots are obtained by setting each linear factor equal to zero. For the factor \\(x-4\\), solve \\(x-4=0\\); adding 4 to both sides gives \\(x=4\\). For the factor \\(x+7\\), solve \\(x+7=0\\); subtracting 7 gives \\(x=-7\\). Therefore the roots are 4 and -7.

Equivalently, the quadratic can be represented, apart from a nonzero constant multiplier, by \\((x-4)(x+7)=0\\). The product is zero when either factor is zero, so these are exactly the two roots. A factor written as \\(x-r\\) has root \\(r\\), while a factor written as \\(x+r\\) has root \\(-r\\). Thus option A is correct. The alternatives with both signs positive or both negative do not satisfy the two given factors.

For a quadratic equation \(x^2+px+25=0\), the sum of the roots is \(-p\). The given roots have sum \(-5+(-5)=-10\). Hence, \(-p=-10\), so \(p=10\). Exam tip: In \(x^2+bx+c=0\), the sum of the roots is \(-b\), not \(b\).

Putting (x=\frac{4}{3}) gives (\frac{16}{3}-\frac{28}{3}+c=0), so (c=4). For fractional roots calculate each term separately.

Using the given coefficients, the equation is \(x^2-16x+64=0\). It factors as \((x-8)^2=0\), so both roots are \(8\). Thus, the equation has equal roots. In an exam, you can also verify this using the discriminant \(D=b^2-4ac\); here, \(D=0\).

A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.

The equation \(x^2+x-20=0\) factors as \((x+5)(x-4)=0\). Hence, \(x=-5\) or \(x=4\), so the roots are 4 and -5. Option B reverses the signs of both roots. Exam tip: Set each linear factor equal to zero to obtain the roots.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\) and their product is \(\alpha\beta=\frac{c}{a}\). Here, \(a=1, b=-8, c=15\), so \(\alpha+\beta=8\) and \(\alpha\beta=15\). Therefore, \(\alpha+\beta+\alpha\beta=8+15=23\). The values 8 and 15 represent only the sum and product respectively, not their total. Exam tip: identify the sum and product of the roots first, then substitute them into the required expression.

For \(ax^2+bx+c=0\), the discriminant \(D=b^2-4ac\) determines root nature. If \(D>0\), roots are distinct real; if \(D=0\), they are equal. In exams, first write the equation in standard form.

The equation \(16x^2-8x+1=0\) can be written as \((4x-1)^2=0\). Hence, \(4x-1=0\), giving the repeated root \(x=\frac{1}{4}\). Exam tip: when a quadratic is a perfect square, its single root is repeated twice; do not confuse it with the coefficient 4 in the factor.

A root is a value that makes the left-hand side of the equation equal to zero. Substituting (x=0) gives a(0)^2+b(0)+c=c, so c=0 is necessary. The condition (a\ne0) only ensures that the equation is genuinely quadratic; it does not imply a=0 or b=0. Exam tip: if zero is a root of a quadratic equation, its constant term must be zero.

A monic quadratic equation has leading coefficient 1. If its roots are \(\alpha\) and \(\beta\), its form is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). Here, the sum of the roots is \(2+7=9\), so the coefficient of \(x\) is \(-9\). Option B results from missing the negative sign. Exam tip: In a monic quadratic, the coefficient of \(x\) is the negative of the sum of the roots.

Factor the quadratic as \(3x^2+2x-8=(3x-4)(x+2)\). Thus, \((3x-4)(x+2)=0\) gives \(x=\frac{4}{3}\) or \(x=-2\). Therefore, option A is correct. In option B, the signs of both roots are incorrect. Exam tip: After factorising, set each factor equal to zero to obtain the roots.

For the quadratic equation \\(x^2+mx+20=0\\), the sum of the roots is \\(-m\\). The given roots have sum \\((-4)+(-5)=-9\\), so \\(-m=-9\\), which gives \\(m=9\\). Therefore, 9 is correct. Exam tip: for \\(x^2+bx+c=0\\), the sum of the roots is \\(-b\\).

Factor the quadratic: \(x^2-5x-24=(x-8)(x+3)\). Hence, \(x=8\) or \(x=-3\). Only 8 is positive, so option A is correct. Exam tip: Always check the signs of both roots before selecting the positive root.

When the discriminant \(D=0\), the two roots of the quadratic equation are equal. If each root is \(r\), then \(r+r=12\), so \(2r=12\) and \(r=6\). Exam tip: When \(D=0\), assume equal roots first and use their sum or product accordingly.

The equation \(x^2-6x-7=0\) factors as \((x-7)(x+1)=0\). Hence, its roots are \(7\) and \(-1\), and their absolute difference is \(|7-(-1)|=8\). Exam tip: while finding the difference between roots, retain the signs of both roots before subtracting.