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ICSE Class X Math 2016 Paper

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Question : 20 of 57
Marks: +1, -0
SECTION - II
Solve the quadratic equation x2−3(x+3)=x^2-3(x+3)= 0 ; Give your answer correct to two significant figures.
Solution:  
Given : x2−3(x+3)=0x^2-3(x+3)=0
x2−3x−9=0x^2-3x-9=0
Comparing x2−3x−9=0x^2-3x-9=0 with ax2+bx+cax^2+bx+c =0=0, we get
We know,
a=1,  b=−3,  c=−9a=1,\; b=-3,\; c=-9
x  =  −b±b2−4ac2ax\;=\;\frac{-b \pm \sqrt{b^2-4ac}}{2a}
  =  −(−3)±(−3)2−4×1×(−9)2×1\;=\;\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times (-9)}}{2 \times 1}
  =  3±9+362  =  3±452\;=\;\frac{3 \pm \sqrt{9+36}}{2}\;=\;\frac{3 \pm \sqrt{45}}{2}
  =  3±352  =  3±3×2.2362\;=\;\frac{3 \pm 3\sqrt{5}}{2}\;=\;\frac{3 \pm 3 \times 2.236}{2}
x  =  3±6.7082x\;=\;\frac{3 \pm 6.708}{2}
x  =  3+6.7082  and  3−6.7082x\;=\;\frac{3+6.708}{2} \; \text{and} \; \frac{3-6.708}{2}
x  =4.85  , and   −1.85x\;=4.85 \; \text{, and } \; -1.85
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