The roots of (x^2+px+36=0) are equal and positive. What is the value of (p)?
Correct answer: A
For equal roots, (p^2-144=0) gives (p=\pm12). For equal positive roots, (-\frac{p}{2}>0), so (p=-12).
Mathematics
Introduction to Quadratic Equations
द्विघात समीकरणों का परिचय
Introduction to Quadratic Equations, part of the Class 10 Mathematics chapter Quadratic Equations, helps students recognise equations of degree two and write them in the standard form ax² + bx + c = 0, where a ≠ 0. Students learn the meaning of coefficients, variables, and constants, identify quadratic equations from examples, and understand how their roots or solutions relate to the equation. The topic builds a foundation for solving quadratic equations by methods such as factorisation and applying the quadratic formula.
Expert · Level 4 · 25 questions
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Expert · Level 4
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- (-12)
- (12)
- (6)
- (-6)
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- (k\neq1)
- (k=1)
- (k\neq2)
- (k=2)
Expert · Level 4
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- (x^2-144=0)
- (x^2+144=0)
- (x^2+12x-144=0)
- (x^2-12x-144=0)
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- (15)
- (-15)
- (9)
- (-9)
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- (504)
- (216)
- (288)
- (-504)
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- -61
- 61
- 49
- -49
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- Contradictory statement
- Quadratic equation
- Linear equation
- Always true statement
Expert · Level 4
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- (x^2-10x+57=0)
- (x^2+10x+57=0)
- (x^2-14x+57=0)
- (x^2-10x-57=0)
Expert · Level 4
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- 0
- \(\frac{2}{3}\)
- \(-\frac{2}{3}\)
- \(\frac{4}{3}\)
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- 28
- 13
- 20
- 32
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- -(\frac{5}{2}) / (-\frac{5}{2})
- (\frac{5}{2})
- (5)
- (-5)
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- \(0\)
- \(7\)
- \(63\)
- \(-49\)
Expert · Level 4
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- \(x^2+12x-112=0\)
- \(x^2+12x-72=0\)
- \(x^2+12x+32=0\)
- \(x^2+12x-144=0\)
Expert · Level 4
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- 225
- 44
- 169
- 196
Expert · Level 4
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- \(3x^2-5x+1=0\)
- \(x(x+2)=3\)
- \((x+1)^2=x^2+4x\)
- \((2x-1)(x+3)=x\)
Expert · Level 4
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- 12
- 6
- 2a
- a+6
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- ( \frac{37}{6} )
- ( \frac{49}{6} )
- ( \frac{31}{6} )
- ( \frac{7}{3} )
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- (3)
- (2)
- (4)
- (5)
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- (6)
- (4)
- (5)
- (8)
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- (4x^2+11x-13=0)
- (4x^2+17x-13=0)
- (4x^2+11x+13=0)
- (4x^2+23x-17=0)
Expert · Level 4
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- (t\neq 8)
- (t\neq -8)
- (t\neq \pm8)
- (t=\pm8)
Expert · Level 4
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- 5x^2-10x-39=0
- 5x^2+10x-39=0
- 5x^2-10x+39=0
- 5x^2-16x-74=0
Expert · Level 4
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- (16x^2+5x-74=0)
- (16x^2-5x-74=0)
- (4x^2+5x-74=0)
- (16x^2+5x+74=0)
Expert · Level 4
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- 9
- 6
- -9
- -6
Expert · Level 4
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- (x^2+15x+56=0)
- (x^2-15x+56=0)
- (x^2+56x+15=0)
- (x^2-56x+15=0)
If (x=-2) is not a root of (kx^2+5x+6=0), what condition must (k) satisfy?
Correct answer: A
Putting (x=-2), the left side becomes (4k-10+6=4k-4). For it not to be a root, (4k-4\neq0), so (k\neq1).
Which quadratic equation has sum of roots (0) and product (-144)?
Correct answer: A
The monic equation is (x^2-(\text{sum})x+\text{product}=0). Using sum (0) and product (-144) gives (x^2-144=0).
If in \(5x^2+kx+45=0\), the product of roots is (-3) times the sum of roots, what is (k)?
Correct answer: A
The product is (9) and the sum is \(-\frac{k}{5}\). From \(9=-3\left(-\frac{k}{5}\right)\), we get \(k=15\).
What is the sum of cubes of the roots of (x^2-6x-16=0)?
Correct answer: A
The sum of roots is (6) and the product is (-16). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=216+288=504).
If the roots of the quadratic equation \(x^2+bx+c=0\) are \(-5\) and \(11\), what is the value of \(b+c\)?
Correct answer: A
For the quadratic equation \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\). Here, the sum is \(-5+11=6\), so \(-b=6\), giving \(b=-6\). Their product is \((-5)(11)=-55\), so \(c=-55\). Therefore, \(b+c=-6-55=-61\). Exam tip: In a monic quadratic, use the root relations sum \(=-b\) and product \(=c\) directly.
If a = 3 is substituted, what type of statement does (a − 3)x² + (a² − 9)x + 11 = 0 become?
Correct answer: A
Substituting a = 3 gives a − 3 = 0 and a² − 9 = 9 − 9 = 0. Therefore, the equation reduces to 11 = 0, which is a false and contradictory statement, so it has no solution. Exam tip: When the coefficients of the variable terms become zero, check whether the remaining constant statement is true or false; a false statement has no solution.
What is the standard form of ((x+5)^2+(x-6)^2=(x+2)^2)?
Correct answer: A
Expanding gives left side (2x^2-2x+61) and right side (x^2+4x+4). Subtracting gives (x^2-6x+57=0).
If \(x=-2\) is a root of the equation \(x^2+(5k-2)x+4k=0\), what is the value of \(k\)?
Correct answer: D
Since \(x=-2\) is a root, substituting it into the equation gives \((-2)^2+(5k-2)(-2)+4k=0\). Therefore, \(4-10k+4+4k=0\), so \(8-6k=0\) and \(k=\frac{4}{3}\). Hence, option D is correct. Exam tip: When a root is given, substitute it directly into the quadratic equation and solve for the parameter.
The roots of the equation \(x^2-8x+5=0\) are \(\alpha\) and \(\beta\). What is the value of \(\alpha+\beta+4\alpha\beta\)?
Correct answer: A
By Vieta’s formulas for \(x^2-8x+5=0\), the sum of the roots is \(\alpha+\beta=8\) and their product is \(\alpha\beta=5\). Therefore, \(\alpha+\beta+4\alpha\beta=8+4(5)=28\). Option 13 results from adding the product only once, so it is incorrect. Exam tip: for \(ax^2+bx+c=0\), the sum of roots is \(-b/a\) and the product is \(c/a\).
If the roots of (x^2+px+q=0) are (5) and (p), what is the value of (p)?
Correct answer: A
The sum of roots is (5+p), and in the equation the sum is (-p). Thus (5+p=-p), giving (p=-\frac{5}{2}).
If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2-16x+63=0\), what is the value of \((\alpha-7)(\beta-7)\)?
Correct answer: A
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.
A right triangle has a base of (x+4) units, a height of (x+8) units, and an area of 72 square units. Which quadratic equation is formed from this information?
Correct answer: A
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 \(\frac{1}{2}\).
If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2-15x+44=0\), what is the value of \((\alpha+\beta)^2\)?
Correct answer: A
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.
After simplification, which of the following equations does not remain a quadratic equation in x?
Correct answer: C
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.
If the equation \(x^2-2ax+a^2-36=0\) is given, what is the difference between its larger and smaller roots?
Correct answer: A
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.
If (\alpha,\beta) are roots of (2x^2-7x+3=0), what is the value of (\frac{\alpha}{\beta}+\frac{\beta}{\alpha})?
Correct answer: A
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}).
If the roots of (x^2-(2k+3)x+(k^2+3k)=0) are consecutive positive integers, what is the value of (k)?
Correct answer: A
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.
If roots of (x^2+px+q=0) are (p+1) and (q-1), and (p+q=5), what is the value of (pq)?
Correct answer: A
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).
What is the standard quadratic form of ((4x-3)(x+5)=2(3x-1))?
Correct answer: A
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).
If ((t^2-64)x^2+(t-8)x+5=0) is a quadratic equation, what is the correct condition on (t)?
Correct answer: C
For the equation to be quadratic, the coefficient of (x^2) must not be (0). Here (t^2-64\neq0), so (t\neq\pm8).
What is the standard form of the equation \(3(x+1)^2+2(x-4)^2=74\)?
Correct answer: A
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 \(3+32-74=-39\)).
What is the standard form with integer coefficients of (\frac{4x^2-3}{5}+\frac{x-2}{4}=3)?
Correct answer: A
Multiplying the whole equation by (20) gives (16x^2-12+5x-10=60). Therefore the standard form is (16x^2+5x-82=0).
If \(x=-6\) is a root of \(2x^2+px-18=0\), what is the value of \(p\)?
Correct answer: A
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.
Which monic quadratic equation has sum of roots (-15) and product (56)?
Correct answer: A
A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-15) gives (x^2+15x+56=0).
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