Roots of a Quadratic Equation

A quadratic equation \(ax^2+bx+c=0\) has no real roots when its discriminant \(D=b^2-4ac\) is less than zero. Here, \(a=1\), \(b=-2\), and \(c=p+3\), so \(D=(-2)^2-4(1)(p+3)=-4(p+2)\). Thus, \(-4(p+2)<0\), which gives \(p>-2\). Remember that \(D=0\) gives one repeated real root, so \(p=-2\) is not included.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Since (\alpha\beta=-\frac{5}{2}) and (\alpha+\beta=-\frac{3}{2}), the value is (\frac{15}{4}).

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=k\), and \(c=16\). Thus, \(b^2-4ac=0\) gives \(k^2-4(1)(16)=0\), so \(k^2-64=0\). Therefore, \(k=8\) or \(k=-8\). For option B, the discriminant would not be zero. Exam tip: For equal roots of a quadratic equation, directly use \(b^2-4ac=0\).

Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as (\pm\sqrt{57}), so (a=\mp\sqrt{57}).

The sum is (\frac{5}{4}) and the product is (\frac{3}{8}). Multiply (x^2-\frac{5}{4}x+\frac{3}{8}=0) by (8).

The equation x^2-5x+6=0 has roots 2 and 3, since it factors as (x-2)(x-3). Therefore α+1 and β+1 are 2 and 3, so α and β are 1 and 2. Their sum is α+β=3 and their product is αβ=2. For x^2+px+q=0, Vieta’s relations give α+β=-p and αβ=q. Thus -p=3, giving p=-3, and q=2. Hence option A is correct. Alternatively, shifting the known roots backward by 1 directly gives the polynomial (x-1)(x-2)=x^2-3x+2, which has the form x^2+px+q. The other options have an incorrect sign or interchange the required parameter values.

A quadratic equation has real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2(k+1)\), and \(c=k^2\), so \(D=b^2-4ac=4(k+1)^2-4k^2=4(2k+1)\). Thus, \(4(2k+1)>0\), which gives \(k>-rac{1}{2}\). Option B is incorrect because at \(k=-rac{1}{2}\), the roots are equal rather than distinct. Exam tip: For ‘real and distinct’ roots, use the strict condition \(D>0\).

By Vieta’s formulas, \(\alpha+\beta=5\) and \(\alpha\beta=1\). Hence, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=5^2-2(1)=25-2=23\). Therefore, the correct answer is 23. In the exam, remember the identity \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\); using only \((\alpha+\beta)^2\) gives 25, which is incorrect.

The monic equation is (x^2-x-6=0). Since the coefficient of (x^2) must be (2), multiply the whole equation by (2).

Use ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta). Since (16=(\alpha+\beta)^2-20), we get ((\alpha+\beta)^2=36) and (m^2=36).

By Vieta’s relations, the sum of the roots is \(10\) and their product is \(k\). The distinct prime numbers that sum to 10 are \(3\) and \(7\); the pair \(5,5\) is excluded because the roots must be distinct. Hence, \(k=3\times7=21\). Exam tip: In \(x^2-Sx+P=0\), the sum of the roots is \(S\) and their product is \(P\).

For \(x^2+2x+c=0\), the sum of the roots is \(-2\), and their product is \(c\). The statement says the product is less than the sum, so \(c<-2\). This condition is already stronger than the real-root requirement. Indeed, the discriminant is \(D=2^2-4c=4-4c\), which is positive whenever \(c<1\), and every value satisfying \(c<-2\) also satisfies that condition.

Thus the complete combined condition is \(c<-2\), which is option A. The inequality must be strict because “less than” does not allow equality. Option B, \(c\le1\), only guarantees real roots and does not guarantee that the product is less than the sum. Therefore the supplied answer and explanation are correct.

Let the roots be \(\alpha\) and \(\beta\). By Vieta’s formula, \(\alpha\beta=m\). For two roots to be reciprocals of each other, \(\alpha\beta=1\); hence, \(m=1\). Indeed, when \(m=1\), the equation becomes \((x-1)^2=0\), so both roots are 1, and each is the reciprocal of the other. The value \(m=0\) is invalid because it gives a zero root, whose reciprocal is undefined. Exam tip: For reciprocal roots, set their product equal to 1.

For equal roots, (k^2-16=0), so (k=\pm4). The equal root is (-\frac{k}{4}), which is negative only when (k=4).

((\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4). Since (\alpha+\beta=4) and (\alpha\beta=2), the value is (14).

Let the roots be \(\alpha\) and \(\beta\). Then \(\alpha+\beta=-6\) and \(\alpha\beta=r\). Using \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\), we get \(20=36-4r\), so \(r=4\). Exam tip: Square the difference of the roots and relate it to their sum and product; the sign of the sum does not affect its square.

Substitute the proposed root into the equation: \(a(0)^2+b(0)+c=0\), which gives \(c=0\). Thus the constant term must be zero. Neither \(b=0\) nor \(a+b=0\) is necessary, while \(a=0\) would make the equation non-quadratic. Exam tip: If zero is a root of a polynomial, its constant term is always zero.

Since \(4x^2-4x+1=(2x-1)^2\), the equation gives \((2x-1)^2=0\), so \(2x-1=0\) and \(x=\frac{1}{2}\). Also, the discriminant \(D=b^2-4ac\) is zero, confirming that the two roots are equal. Therefore, both roots are \(\frac{1}{2},\frac{1}{2}\). Exam tip: A quadratic that is a perfect square, or has \(D=0\), has equal roots.

For equal roots, (s^2-36=0), so (s=\pm6). The equal root is (\frac{s}{2}), which is positive when (s=6).

For a quadratic equation (ax^2+bx+c=0) to have equal roots, its discriminant (D=b^2-4ac) must be zero. Here, a=k-2, b=4, c=1, so D=16-4(k-2)=0, giving k=6. Note that when k=2, the equation becomes linear rather than quadratic, so it cannot represent equal roots of a quadratic equation. Exam tip: Set the discriminant equal to zero whenever a quadratic is stated to have equal roots.

((\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4). Since (\alpha+\beta=6) and (\alpha\beta=8), the value is (0).

The governing concept is transformation of roots. First, x² − 6x + 5 = (x − 1)(x − 5), so the original roots are α = 1 and β = 5. Applying the transformation 3t − 2 gives new roots 3(1) − 2 = 1 and 3(5) − 2 = 13. The monic quadratic with roots 1 and 13 is (x − 1)(x − 13) = x² − 14x + 13, so option A is correct. Using Vieta directly gives the same result: the new sum is 3(α + β) − 4 = 3(6) − 4 = 14, and the new product is (3α − 2)(3β − 2) = 9αβ − 6(α + β) + 4 = 45 − 36 + 4 = 13. The other options have an incorrect sum, product, or sign.

The governing concept is testing a proposed root by substitution and then using factorisation. For f(x)=x^2−(a+4)x+4a, substitute x=4: f(4)=16−4(a+4)+4a=16−4a−16+4a=0. The expression is zero for every value of a, so 4 is always a root. The complete factorisation is x^2−(a+4)x+4a=(x−4)(x−a), which shows that the other root is a. Option B is not generally true because a+4 does not normally satisfy the equation. The roots coincide only when a=4, and for every real a the two roots 4 and a are real. Consequently, C and D are false, leaving option A as the only universally valid statement.

The governing concept is recognising a difference of squares. Rewrite the quadratic as x^2−2px+p^2−9=(x−p)^2−3^2. Therefore it factors as (x−p−3)(x−p+3)=0, giving roots p+3 and p−3. The larger root minus the smaller root is (p+3)−(p−3)=6, independent of p, so option A is correct. Vieta’s relations confirm that the sum is 2p, not a fixed 6, and the product is p^2−9, not a fixed 9. The roots are distinct because their difference is 6, so they cannot always be equal. This reasoning remains valid for every real value of p, and no additional restriction is needed for the stated constant difference.

Use the relationship between roots and coefficients, or factor directly. The expression x^2−(2a+1)x+a(a+1) equals (x−a)(x−a−1), because the sum of a and a+1 is 2a+1 and their product is a(a+1). Therefore the equation has roots x=a and x=a+1, making option A correct. Option B incorrectly treats the coefficient of x as a root. The roots are not always equal; they differ by 1 for every a. Since both roots are real whenever a is real, option D is also incorrect. The discriminant is 1, confirming two distinct real roots.