Question : Which of the following equations has 7 as a root?
Option 1: $3x^2-6x+2=0$
Option 2: $x^2-9x+14=0$
Option 3: $x^2-7x+10=0$
Option 4: $x^2+3x-12=0$
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $x^2-9x+14=0$
Solution : Given: Option 1: $3x^2-6x+2=0$ Using the quadratic formula, we get, $⇒x=\frac{6\pm\sqrt{(-6)^2-4.3.2}}{2.3}$ So, $x=\frac{\sqrt3+3}{3}$ or $x=\frac{-\sqrt{3}+3}{3}$ So, the value of $x\neq7$. Option 2: $x^2-9x+14=0$ $⇒x^2-7x-2x+14=0$ $⇒(x-7)(x-2)=0$ $\therefore x = 7$ or $x = 2$ Here, the value of $x=7$. Hence, the correct answer is $x^2-9x+14=0$.
Candidates can download this e-book to give a boost to thier preparation.
Application | Eligibility | Admit Card | Answer Key | Preparation Tips | Result | Cutoff
Question : Which of the following equations has equal roots?
Option 1: $3x^{2}-6x+2=0$
Option 2: $3x^{2}-6x+3=0$
Option 3: $x^{2}-8x+8=0$
Option 4: $4x^{2}-8x+2=0$
Question : Simplify the following expression. $\{[(x-5)(x-1)]-[(9 x-5)(9x-1)]\} \div 16x$
Option 1: $2x(5x-3)$
Option 2: $-(5x-3)$
Option 3: $x(5x-3)$
Option 4: $-6x(5x-3)$
Question : If $x+\frac{2}{x}=1$, then the value of $\frac{x^2+7x+2}{x^2+13x+2}$ is:
Option 1: $\frac{5}{7}$
Option 2: $\frac{3}{7}$
Option 3: $\frac{4}{7}$
Option 4: $\frac{2}{7}$
Question : If $x+\frac{1}{x}=-2$, then what is the value of $x^7+x^{-7}+x^2+x^{-2} ?(\mathrm{x}<0)$
Option 1: 4
Option 2: 2
Option 3: 1
Option 4: 0
Question : $9x^2+25 - 30x$ can be expressed as the square of:
Option 1: $-3x - 5$
Option 2: $3x+5$
Option 3: $3x-5$
Option 4: $3x^2-25$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile