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Maths 1 Marks Question

Quadratic Equations · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 19 questions.

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19 questions · timed · auto-graded

Question 11 Mark
Find the values of k so that the quadratic equation $9 x^2-3 k x+k=0$ has equal roots.
Answer
Given equation is $9 x^2-3 k x+k=0$
Comparing with $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$
$a=9, b=-3 k, c=k$
For real and equal roots, we must have
$ \text { Discriminant, } D=0 $
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow(-3 k)^2-4 \times 1 \times k=0 $
$ \Rightarrow 9 k^2-36 k=0 $
$ \Rightarrow 9 k(k-4)=0 $
$ \Rightarrow 9 k=0 \text { or } k-4=0 $
$ \Rightarrow k=0 \text { 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 $3 x^2-10 x+k=0$  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 $3 x^2-10 x+k=0$
On comparing with $a x^2+b x+c=0$, 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 $x^2+6 x+9=0$.
Answer
Given equation is $x^2+6 x+9=0$
Substituting $x = -3$ in $L.H.S$. of above equation, we get
$L.H.S.$ $= (-3)^2+ 6 × (-3) + 9$
$= 9 - 18 + 9$
$= 18 - 18$
$=0$
$= R.H.S.$
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 8 x-x^2=15 $
$ \Rightarrow x^2-8 x+15=0 $
$ \Rightarrow x^2-3 x-5 x+15=0 $
$ \Rightarrow x(x-3)-5(x-3)=0 $
$ \Rightarrow(x-3)(x-5)=0 $
$ \Rightarrow x-3=0 \text { or } x-5=0 $
$ \Rightarrow x=3 \text { 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
$x^2- x + 3 = 0$ 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:
$ 48 x^2-13 x-1=0 $
Answer
$ 48 x^2-13 x-1=0 $
$ \Rightarrow 48 x^2-16 x+3 x-1=0 $
$ \Rightarrow 16 x(3 x-1)+1(3 x-1)=0 $
$ \Rightarrow(3 x-1)(16 x+1)=0 $
$ \Rightarrow 3 x-1=0 \text { or } 16 x+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
$\text { Since the roots of the equation } 2 x^2+8 x+k=0 \text { are equal, }$
$ \text { Discriminant, } \mathrm{D}=0 $
$ \Rightarrow \mathrm{~b}^2-4 \mathrm{ac}=0 $
$ \Rightarrow(8)^2-4 \times 2 \times \mathrm{k}=0 $
$ \Rightarrow 64-8 \mathrm{k}=0 $
$ \Rightarrow 8 \mathrm{k}=64 $
$ \Rightarrow \mathrm{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 $3 x^2+13 x+14=0$.
Answer
Given equation is $3 x^2+13 x+14=0$
Substituting $x = -2$ in $L.H.S$. of above equation, we get
$L.H.S.$ $= 3(-2)^2+ 13 × (-2) + 14$
$= 12 - 26 + 14$
$= 26 - 26$
$= 0$
$= R.H.S.$
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 $px(x - 2) + 6 = 0$
$\Rightarrow p x^2-2 p x+6=0$
Comparing with $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$, we have
$a=p, b=-2 p, c=6$
Since the roots are equal, we have
Discriminant, $D=0$
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow(-2 p)^2-4 \times p \times 6=0 $
$ \Rightarrow 4 p^2-24 p=0 $
$ \Rightarrow 4 p(p-6)=0 $
$ \Rightarrow 4 p=0 \text { or } p-6=0 $
$ \Rightarrow p=6(\text { since } p \neq 0)$
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Question 191 Mark
Find the roots of the quadratic equation $ 2 x^2-x-6=0 $
Answer
$ 2 x^2-x-6=0 $
$ \Rightarrow 2 x^2-4 x+3 x-6=0 $
$ \Rightarrow 2 x(x-2)+3(x-2)=0 $
$ \Rightarrow(x-2)(2 x+3)=0 $
$ \Rightarrow x-2=0 \text { or } 2 x+3=0$
$ ⇒ x = 2$ or $\text{x}=-\frac{3}{2}$
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