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Maths 1 Marks Question

Quadratic Equations · Maharashtra Board STD 10 (MSBSHSE - English Medium). 19 questions.

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Question 11 Mark
Find the values of k so that the quadratic equation $9x^2- 3kx + k = 0$ has equal roots.
Answer
Given equation is $9x^2- 3kx + k = 0$
Comparing with $ax^2+ bx + c = 0$
$a = 9, b = -3k, c = k$
For real and equal roots, we must have
Discriminant, $D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow (-3k)^2 - 4 \times 1 \times k = 0$
$\Rightarrow 9k^2 - 36k = 0$
$\Rightarrow 9k(k - 4) = 0$
$\Rightarrow 9k = 0$ or $k - 4 = 0$
$\Rightarrow k = 0$ or $k = 4$
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Question 21 Mark
If $1$ is a root of the equation $ay^2 + ay + 3 = 0$ and $y^2 + y + b = 0$ then find the value of ab.
Answer
Since $1$ is a root of the equation $ay^2 + ay + 3 = 0$, we have
$a(1)^2 + a(1) + 3 = 0$
$\Rightarrow a + a + 3 = 0$
$\Rightarrow 2a = -3$
$\Rightarrow\text{a}=\frac{-3}{2}$
Also, $1$ is a root of the equation $y^2 + y + b = 0$
$\Rightarrow (1)^2 + 1 + b = 0$
$\Rightarrow 2 + b = 0$
$\Rightarrow b = -2$
Hence, $\text{ab}=\frac{-3}{2}\times(-2)=3$
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Question 31 Mark
If one root of the quadratic equation $3x^2 - 10x + k = 0$ is reciprocal of the other, find the value of $k$.
Answer
Let one root of the given equation be $\alpha.$
Then, its root will be $\frac{1}{\alpha}.$
Given equation is $3x^2 - 10x + k = 0$
On comparing with $ax^2 + bx + c = 0$, we have
$a = 3, b = -10, c = k$
Now,
Product of the roots $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\times\frac{1}{\alpha}=\frac{\text{k}}3{}$
$\Rightarrow\text{k}=3$
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Question 41 Mark
Show that $x = -3$ is a solution of $x^2 + 6x + 9 = 0$.
Answer
Given equation is $x^2 + 6x + 9 = 0$
Substituting $x = -3$ in L.H.S. of above equation, we get
L.H.S. $= (-3)^2 + 6 \times (-3) + 9$
$= 9 - 18 + 9$
$= 18 - 18$
$= 0$
= R.H.S.
Hence, $x = -3$ is a solution of the given equation.
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Question 51 Mark
The sum of two natural numbers is $8$ and their product is $15$. Find the numbers.
Answer
Let the required natural numbers be $x$ and $(8 - x)$. Then,
we have $x(8 - x) = 15 $
$\Rightarrow 8x - x^2 = 15 $
$\Rightarrow x^2 - 8x + 15 = 0 $
$\Rightarrow x^2 - 3x - 5x + 15 = 0 $
$\Rightarrow x(x - 3) - 5(x - 3) = 0$
$\Rightarrow (x - 3)(x - 5) = 0$
$\Rightarrow x - 3 = 0$ or $x - 5 = 0$
$\Rightarrow x = 3$ or $x = 5$
Hence, the required numbers are $3$ and $5$.
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Question 61 Mark
If the quadratic equation $\text{px}^2-2\sqrt5\text{px}+15=0$ has equal roots then find the value of $p$.
Answer
Since the roots of the equation $\text{px}^2-2\sqrt5\text{px}+15=0$ are equal,
Discriminant, $D = 0$
$\Rightarrow\big(-2\sqrt5\text{p}\big)^2-4\times\text{p}\times15=0$
$\Rightarrow 20p^2 - 60p = 0$
$\Rightarrow 20p(p - 3) = 0$
$\Rightarrow 20p = 0$ or $p - 3 = 0$
$\Rightarrow p = 3$ $($Since $\text{p}\neq0)$
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Question 71 Mark
The following are quadratic equations in x?
$x^2 - x + 3 = 0$
Answer
$x^2 - x + 3$ is a quadratic polynomial.
$x^2 - x + 3 = 0$ is a quadratic equation.
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Question 81 Mark
Solve the following quadratic equation:$48x^2 - 13x - 1 = 0$
Answer
$48x^2 - 13x - 1 = 0$
$\Rightarrow 48x^2 - 16x + 3x - 1 = 0$
$\Rightarrow 16x(3x - 1) + 1(3x - 1) = 0$
$\Rightarrow (3x - 1)(16x + 1) = 0$
$\Rightarrow 3x - 1 = 0$ or $16x + 1 = 0$
$\Rightarrow\text{x}=\frac{1}{3}$ or $\text{x}=\frac{-1}{16}$
Hence, $\frac{-1}{16}$ and $\frac{1}{3}$ are the roots of the equation $48x^2 - 13x - 1 = 0$
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Question 91 Mark
If $\text{x}=\frac{-1}{2}$ is a solution of the quadratic equation $3x^2 + 2kx - 3 = 0$, find the value of $k$.
Answer
Since $\text{x}=\frac{-1}{2}$ is a solution of equation $3x^2 + 2kx - 3 = 0,$
$3\Big(\frac{-1}{2}\Big)+\text{2k}\times\Big(\frac{-1}{2}\Big)-3=0$
$\Rightarrow3\times\frac{1}{4}-\text{k}-3=0$
$\Rightarrow\text{k}=\frac{3}{4}-3$
$\Rightarrow\text{k}=\frac{3-12}{4}$
$\Rightarrow\text{k}=\frac{-9}{4}$
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Question 101 Mark
The following are quadratic equations in x?
$\frac{1}{3}\text{x}^2+\frac{1}{5}\text{x}-2=0$
Answer
$\frac{1}{3}\text{x}^2+\frac{1}{5}\text{x}-2=0$
$\Rightarrow\text{5x}^2+\text{3x}-30=0$
Clearly, $\text{5x}^2+\text{3x}-30=0$ is a qudratic equation.
$\therefore\ \frac{1}{3}\text{x}^2+\frac{1}{5}\text{x}-2=0$ is a quadratic equation.
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Question 111 Mark
Find the values of k so that the quadratic equation $x^2- 4kx + k = 0$ has equal roots.
Answer
Given equation is $x^2- 4kx + k = 0$
Comparing with $ax^2+ bx + c = 0$
$a = 1, b = -4k, c = k$
Since the roots are equal, we have
Discriminant, $D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow (-4k)^2 - 4 \times 1 \times k = 0$
$\Rightarrow 16k^2 - 4k = 0$
$\Rightarrow 4k(4k - 1) = 0$
$\Rightarrow 4k = 0$ or $4k - 1 = 0$
$\Rightarrow k = 0$ or $\text{k}=\frac{1}{4}$
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Question 121 Mark
The following are quadratic equations in x?
$\sqrt{2\text{x}^2}+7\text{x}+5\sqrt{2}=0$
Answer
$\sqrt{2\text{x}^2}+7\text{x}+5\sqrt{2}$ is a qudratic polynomial.
$\therefore\ \sqrt{2\text{x}^2}+7\text{x}+5\sqrt{2}=0$ is a quadratic equation.
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Question 131 Mark
Find the solution of the quadratic equation $3\sqrt3\text{x}^2+\text{10x}+\sqrt3=0.$
Answer
$3\sqrt3\text{x}^2+\text{10x}+\sqrt3=0$ $\Rightarrow3\sqrt3\text{x}^2+\text{9x}+\text{x}+\sqrt3=0$ $\Rightarrow3\sqrt3\big(\text{x}+\sqrt3\big)+1\big(\text{x}+\sqrt3\big)=0$ $\Rightarrow\big(\text{x}+\sqrt3\big)\big(3\sqrt3+1\big)=0$ $\Rightarrow\text{x}+\sqrt3=0$ or $3\sqrt3+1=0$ $\Rightarrow\text{x}=-\sqrt3$ or $\text{x}=-\frac{1}{3\sqrt3}$
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Question 141 Mark
If the roots of the quadratic equation $2 x^2+8 x+k=0$ are equal then find the value of $k$.
Answer
Since the roots of the equation $2 x ^2+8 x + k =0$ are equal,
Discriminant, $D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow (8)^2 - 4 \times 2 \times k = 0$
$\Rightarrow 64 - 8k = 0$
$\Rightarrow 8k = 64$
$\Rightarrow k = 8$
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Question 151 Mark
The following are quadratic equations in x?
$\text{2x}^2+\frac{5}{2}\text{x}-\sqrt{3}=0$
Answer
$\text{2x}^2+\frac{5}{2}\text{x}-\sqrt{3}=0$
$\Rightarrow\text{4x}^2+\text{5x}-2\sqrt3=0$
Clearly, $\text{4x}^2+\text{5x}-2\sqrt3$ is a qudratic polynomial.
$\therefore\ \text{2x}^2+\frac{5}{2}\text{x}-\sqrt{3}=0$ is a quadratic equation.
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Question 161 Mark
If one zero of the polynomial $x^2 - 4x + 1$ is $\big(2+\sqrt3\big),$ write the other zero.
Answer
Given equation is $x^2 - 4x + 1$Let the other root be $\alpha.$
Sum of the roots $=\frac{-(-4)}{1}$
$\Rightarrow\alpha+\big(2+\sqrt3\big)=4$
$\Rightarrow\alpha=4-2-\sqrt3$
$\Rightarrow\alpha=2-\sqrt3$
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Question 171 Mark
Show that $x = -2$ is a solution of $3x^2 + 13x + 14 = 0.$
Answer
Given equation is $3x^2 + 13x + 14 = 0$
Substituting x = -2 in L.H.S. of above equation, we get
$L.H.S. = 3(-2)^2 + 13 \times (-2) + 14$
$= 12 - 26 + 14$
$= 26 - 26$
$= 0$
$= R.H.S.$
Hence, $x = -2$ is a solution of the given equation.
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Question 181 Mark
If the roots of the quadratic equation $px(x - 2) + 6 = 0$ are equal, find the value of $p$.
Answer
Given equation is$ px(x - 2) + 6 = 0$
$\Rightarrow px^2 - 2px + 6 = 0$
Comparing with $ax^2 + bx + c = 0,$ we have
$a = p, b = -2p, c = 6$
Since the roots are equal, we have
Discriminant, $D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow (-2p)^2 - 4 \times p \times 6 = 0$
$\Rightarrow 4p^2 - 24p = 0$
$\Rightarrow 4p(p - 6) = 0$
$\Rightarrow 4p = 0 or p - 6 = 0$
$\Rightarrow p = 6$ $($since $\text{p}\neq0)$
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Question 191 Mark
Find the roots of the quadratic equation $2x^2 − x − 6 = 0$
Answer
$2x^2 − x − 6 = 0$
$\Rightarrow 2x^2 - 4x + 3x - 6 = 0$
$\Rightarrow 2x(x - 2) + 3(x - 2) = 0$
$\Rightarrow (x - 2)(2x + 3) = 0$
$\Rightarrow x - 2 = 0 or 2x + 3 = 0$
$\Rightarrow x = 2 $or $\text{x}=-\frac{3}{2}$
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